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CyC2018
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@ -2769,52 +2769,29 @@ Explanation: The array can be partitioned as [1, 5, 5] and [11].
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可以看成一个背包大小为 sum/2 的 0-1 背包问题。
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```java
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public boolean canPartition(int[] nums) {
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int sum = 0;
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for (int num : nums) sum += num;
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if (sum % 2 != 0) return false;
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int W = sum / 2;
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boolean[] dp = new boolean[W + 1];
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dp[0] = true;
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for (int num : nums) { // 0-1 背包一个物品只能用一次
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for (int i = W; i >= 0; i--) { // 从后往前，先计算 dp[i] 再计算 dp[i-num]
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if (num <= i) {
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dp[i] = dp[i] || dp[i - num];
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}
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}
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}
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return dp[W];
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}
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```
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**字符串按单词列表分割**
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[139. Word Break (Medium)](https://leetcode.com/problems/word-break/description/)
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```html
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s = "leetcode",
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dict = ["leet", "code"].
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Return true because "leetcode" can be segmented as "leet code".
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```
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这是一个完全背包问题，和 0-1 背包不同的是，完全背包中物品可以使用多次。在这一题当中，词典中的单词可以被使用多次。
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0-1 背包和完全背包在实现上的不同之处是，0-1 背包对物品的迭代是在最外层，而完全背包对物品的迭代是在最里层。
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```java
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public boolean wordBreak(String s, List<String> wordDict) {
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int n = s.length();
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boolean[] dp = new boolean[n + 1];
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public boolean canPartition(int[] nums) {
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int sum = computeArraySum(nums);
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if (sum % 2 != 0) {
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return false;
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}
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int W = sum / 2;
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boolean[] dp = new boolean[W + 1];
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dp[0] = true;
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for (int i = 1; i <= n; i++) {
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for (String word : wordDict) { // 每个单词可以使用多次
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int len = word.length();
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if (len <= i && word.equals(s.substring(i - len, i))) {
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dp[i] = dp[i] || dp[i - len];
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}
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Arrays.sort(nums);
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for (int num : nums) { // 0-1 背包一个物品只能用一次
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for (int i = W; i >= num; i--) { // 从后往前，先计算 dp[i] 再计算 dp[i-num]
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dp[i] = dp[i] || dp[i - num];
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}
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}
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return dp[n];
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return dp[W];
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}
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private int computeArraySum(int[] nums) {
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int sum = 0;
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for (int num : nums) {
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sum += num;
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}
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return sum;
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}
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```
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@ -2850,21 +2827,29 @@ sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)
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```java
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public int findTargetSumWays(int[] nums, int S) {
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int sum = 0;
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for (int num : nums) sum += num;
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if (sum < S || (sum + S) % 2 == 1) return 0;
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int sum = computeArraySum(nums);
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if (sum < S || (sum + S) % 2 == 1) {
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return 0;
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}
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int W = (sum + S) / 2;
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int[] dp = new int[W + 1];
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dp[0] = 1;
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Arrays.sort(nums);
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for (int num : nums) {
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for (int i = W; i >= 0; i--) {
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if (num <= i) {
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dp[i] = dp[i] + dp[i - num];
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}
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for (int i = W; i >= num; i--) {
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dp[i] = dp[i] + dp[i - num];
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}
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}
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return dp[W];
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}
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private int computeArraySum(int[] nums) {
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int sum = 0;
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for (int num : nums) {
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sum += num;
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}
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return sum;
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}
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```
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DFS 解法：
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@ -2875,8 +2860,42 @@ public int findTargetSumWays(int[] nums, int S) {
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}
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private int findTargetSumWays(int[] nums, int start, int S) {
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if (start == nums.length) return S == 0 ? 1 : 0;
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return findTargetSumWays(nums, start + 1, S + nums[start]) + findTargetSumWays(nums, start + 1, S - nums[start]);
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if (start == nums.length) {
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return S == 0 ? 1 : 0;
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}
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return findTargetSumWays(nums, start + 1, S + nums[start])
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+ findTargetSumWays(nums, start + 1, S - nums[start]);
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}
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```
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**字符串按单词列表分割**
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[139. Word Break (Medium)](https://leetcode.com/problems/word-break/description/)
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```html
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s = "leetcode",
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dict = ["leet", "code"].
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Return true because "leetcode" can be segmented as "leet code".
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```
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dict 中的单词没有使用次数的限制，因此这是一个完全背包问题。
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0-1 背包和完全背包在实现上的不同之处是，0-1 背包对物品的迭代是在最外层，而完全背包对物品的迭代是在最里层。
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```java
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public boolean wordBreak(String s, List<String> wordDict) {
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int n = s.length();
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boolean[] dp = new boolean[n + 1];
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dp[0] = true;
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for (int i = 1; i <= n; i++) {
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for (String word : wordDict) { // 完全一个物品可以使用多次
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int len = word.length();
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if (len <= i && word.equals(s.substring(i - len, i))) {
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dp[i] = dp[i] || dp[i - len];
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}
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}
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}
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return dp[n];
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}
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```
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@ -2895,13 +2914,18 @@ Explanation: There are totally 4 strings can be formed by the using of 5 0s and
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```java
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public int findMaxForm(String[] strs, int m, int n) {
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if (strs == null || strs.length == 0) return 0;
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if (strs == null || strs.length == 0) {
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return 0;
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}
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int[][] dp = new int[m + 1][n + 1];
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for (String s : strs) { // 每个字符串只能用一次
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for (String s : strs) { // 每个字符串只能用一次
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int ones = 0, zeros = 0;
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for (char c : s.toCharArray()) {
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if (c == '0') zeros++;
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else ones++;
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if (c == '0') {
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zeros++;
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} else {
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ones++;
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}
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}
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for (int i = m; i >= zeros; i--) {
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for (int j = n; j >= ones; j--) {
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@ -2913,7 +2937,7 @@ public int findMaxForm(String[] strs, int m, int n) {
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}
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```
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**找零钱**
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**找零钱的方法数**
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[322. Coin Change (Medium)](https://leetcode.com/problems/coin-change/description/)
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@ -2929,22 +2953,27 @@ return -1.
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题目描述：给一些面额的硬币，要求用这些硬币来组成给定面额的钱数，并且使得硬币数量最少。硬币可以重复使用。
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这是一个完全背包问题，完全背包问题和 0-1 背包问题在实现上的区别在于，0-1 背包遍历物品的循环在外侧，而完全背包问题遍历物品的循环在内侧，在内侧体现出物品可以使用多次。
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- 物品：硬币
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- 物品大小：面额
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- 物品价值：数量
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因为硬币可以重复使用，因此这是一个完全背包问题。
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```java
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public int coinChange(int[] coins, int amount) {
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if (coins == null || coins.length == 0) return 0;
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int[] dp = new int[amount + 1];
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Arrays.fill(dp, amount + 1);
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dp[0] = 0;
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if (coins == null || coins.length == 0) {
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return 0;
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}
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int[] minimum = new int[amount + 1];
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Arrays.fill(minimum, amount + 1);
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minimum[0] = 0;
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Arrays.sort(coins);
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for (int i = 1; i <= amount; i++) {
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for (int c : coins) { // 硬币可以使用多次
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if (c <= i) {
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dp[i] = Math.min(dp[i], dp[i - c] + 1);
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}
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for (int j = 0; j < coins.length && coins[j] <= i; j++) {
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minimum[i] = Math.min(minimum[i], minimum[i - coins[j]] + 1);
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}
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}
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return dp[amount] > amount ? -1 : dp[amount];
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return minimum[amount] > amount ? -1 : minimum[amount];
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}
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```
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@ -2974,17 +3003,18 @@ Therefore the output is 7.
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```java
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public int combinationSum4(int[] nums, int target) {
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if (nums == null || nums.length == 0) return 0;
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int[] dp = new int[target + 1];
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dp[0] = 1;
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if (nums == null || nums.length == 0) {
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return 0;
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}
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int[] maximum = new int[target + 1];
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maximum[0] = 1;
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Arrays.sort(nums);
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for (int i = 1; i <= target; i++) {
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for (int num : nums) {
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if (num <= i) {
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dp[i] += dp[i - num];
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}
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for (int j = 0; j < nums.length && nums[j] <= i; j++) {
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maximum[i] += maximum[i - nums[j]];
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}
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}
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return dp[target];
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return maximum[target];
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}
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```
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@ -2992,31 +3022,27 @@ public int combinationSum4(int[] nums, int target) {
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[188. Best Time to Buy and Sell Stock IV (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/)
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```html
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dp[i, j] = max(dp[i, j-1], prices[j] - prices[jj] + dp[i-1, jj]) { jj in range of [0, j-1] }
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= max(dp[i, j-1], prices[j] + max(dp[i-1, jj] - prices[jj]))
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```
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```java
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public int maxProfit(int k, int[] prices) {
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int n = prices.length;
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if (k >= n/2) { // 这种情况下该问题退化为普通的股票交易问题
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int maxPro = 0;
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if (k >= n / 2) { // 这种情况下该问题退化为普通的股票交易问题
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int maxProfit = 0;
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for (int i = 1; i < n; i++) {
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if (prices[i] > prices[i-1])
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maxPro += prices[i] - prices[i-1];
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if (prices[i] > prices[i - 1]) {
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maxProfit += prices[i] - prices[i - 1];
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}
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}
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return maxPro;
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return maxProfit;
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}
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int[][] dp = new int[k + 1][n];
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int[][] maxProfit = new int[k + 1][n];
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for (int i = 1; i <= k; i++) {
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int localMax = dp[i - 1][0] - prices[0];
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int localMax = maxProfit[i - 1][0] - prices[0];
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for (int j = 1; j < n; j++) {
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dp[i][j] = Math.max(dp[i][j - 1], prices[j] + localMax);
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localMax = Math.max(localMax, dp[i - 1][j] - prices[j]);
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maxProfit[i][j] = Math.max(maxProfit[i][j - 1], prices[j] + localMax);
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localMax = Math.max(localMax, maxProfit[i - 1][j] - prices[j]);
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}
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}
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return dp[k][n - 1];
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return maxProfit[k][n - 1];
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}
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```
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@ -3029,10 +3055,18 @@ public int maxProfit(int[] prices) {
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int firstBuy = Integer.MIN_VALUE, firstSell = 0;
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int secondBuy = Integer.MIN_VALUE, secondSell = 0;
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for (int curPrice : prices) {
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if (firstBuy < -curPrice) firstBuy = -curPrice;
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if (firstSell < firstBuy + curPrice) firstSell = firstBuy + curPrice;
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if (secondBuy < firstSell - curPrice) secondBuy = firstSell - curPrice;
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if (secondSell < secondBuy + curPrice) secondSell = secondBuy + curPrice;
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if (firstBuy < -curPrice) {
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firstBuy = -curPrice;
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}
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if (firstSell < firstBuy + curPrice) {
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firstSell = firstBuy + curPrice;
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}
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if (secondBuy < firstSell - curPrice) {
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secondBuy = firstSell - curPrice;
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}
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if (secondSell < secondBuy + curPrice) {
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secondSell = secondBuy + curPrice;
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}
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}
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return secondSell;
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}
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