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CyC2018 2019-03-05 17:28:49 +08:00
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docs/notes/Leetcode 题解.md
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@ -3199,31 +3199,34 @@ return -1.
- 物品大小：面额 - 物品大小：面额
- 物品价值：数量 - 物品价值：数量
因为硬币可以重复使用，因此这是一个完全背包问题。完全背包只需要将01背包中逆序遍历bp数组改为正序遍历即可。 因为硬币可以重复使用，因此这是一个完全背包问题。完全背包只需要将 0-1 背包中逆序遍历 dp 数组改为正序遍历即可。
```java ```java
public int coinChange(int[] coins, int amount) { public int coinChange(int[] coins, int amount) {
if(amount == 0) return 0; if (amount == 0 || coins == null || coins.length == 0) {
int[] bp = new int[amount + 1]; return 0;
}
int[] dp = new int[amount + 1];
for (int coin : coins) { for (int coin : coins) {
for (int i = coin; i <= amount; i++) { //将逆序遍历改为正序遍历 for (int i = coin; i <= amount; i++) { //将逆序遍历改为正序遍历
if(i == coin) if (i == coin) {
bp[i] = 1; dp[i] = 1;
else if(bp[i] == 0 && bp[i - coin] != 0) } else if (dp[i] == 0 && dp[i - coin] != 0) {
bp[i] = bp[i - coin] + 1; dp[i] = dp[i - coin] + 1;
else if(bp[i - coin] != 0) } else if (dp[i - coin] != 0) {
bp[i] = Math.min(bp[i], bp[i - coin] + 1); dp[i] = Math.min(dp[i], dp[i - coin] + 1);
} }
} }
return bp[amount] == 0 ? -1 : bp[amount]; }
return dp[amount] == 0 ? -1 : dp[amount];
} }
``` ```
**找零钱的硬币数组合** **找零钱的硬币数组合**
[518. Coin Change 2 (Medium)](https://leetcode.com/problems/coin-change-2/description/) [518\. Coin Change 2 (Medium)](https://leetcode.com/problems/coin-change-2/description/)
```html ```text-html-basic
Input: amount = 5, coins = [1, 2, 5] Input: amount = 5, coins = [1, 2, 5]
Output: 4 Output: 4
Explanation: there are four ways to make up the amount: Explanation: there are four ways to make up the amount:
@ -3237,6 +3240,9 @@ Explanation: there are four ways to make up the amount:
```java ```java
public int change(int amount, int[] coins) { public int change(int amount, int[] coins) {
if (amount == 0 || coins == null || coins.length == 0) {
return 0;
}
int[] dp = new int[amount + 1]; int[] dp = new int[amount + 1];
dp[0] = 1; dp[0] = 1;
for (int coin : coins) { for (int coin : coins) {
@ -3260,7 +3266,7 @@ Return true because "leetcode" can be segmented as "leet code".
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。该问题涉及到字典中单词的使用顺序，因此可理解为涉及顺序的完全背包问题。 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。该问题涉及到字典中单词的使用顺序，因此可理解为涉及顺序的完全背包问题。
求解该类型问题时，应调整两次循环的顺序，0-1 背包对物品的迭代是在最外层，而涉及顺序的完全背包问题对物品的迭代是在最里层。 求解顺序的完全背包问题时，对物品的迭代应该放在最里层。
```java ```java
public boolean wordBreak(String s, List<String> wordDict) { public boolean wordBreak(String s, List<String> wordDict) {
@ -3268,7 +3274,7 @@ public boolean wordBreak(String s, List<String> wordDict) {
boolean[] dp = new boolean[n + 1]; boolean[] dp = new boolean[n + 1];
dp[0] = true; dp[0] = true;
for (int i = 1; i <= n; i++) { for (int i = 1; i <= n; i++) {
for (String word : wordDict) { // 完全一个物品可以使用多次 for (String word : wordDict) { // 对物品的迭代应该放在最里层
int len = word.length(); int len = word.length();
if (len <= i && word.equals(s.substring(i - len, i))) { if (len <= i && word.equals(s.substring(i - len, i))) {
dp[i] = dp[i] || dp[i - len]; dp[i] = dp[i] || dp[i - len];