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CyC2018 2018-06-14 10:18:14 +08:00
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notes/Leetcode 题解.md
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@ -3116,7 +3116,9 @@ the contiguous subarray [4,-1,2,1] has the largest sum = 6.
```java
public int maxSubArray(int[] nums) {
if (nums == null || nums.length == 0) return 0;
if (nums == null || nums.length == 0) {
return 0;
}
int preSum = nums[0];
int maxSum = preSum;
for (int i = 1; i < nums.length; i++) {
@ -3142,7 +3144,9 @@ dp[i] 表示以 A[i] 为结尾的等差递增子区间的个数。
```java
public int numberOfArithmeticSlices(int[] A) {
if (A == null || A.length == 0) return 0;
if (A == null || A.length == 0) {
return 0;
}
int n = A.length;
int[] dp = new int[n];
for (int i = 2; i < n; i++) {
@ -3150,9 +3154,11 @@ public int numberOfArithmeticSlices(int[] A) {
dp[i] = dp[i - 1] + 1;
}
}
int ret = 0;
for (int cnt : dp) ret += cnt;
return ret;
int total = 0;
for (int cnt : dp) {
total += cnt;
}
return total;
}
```
@ -3176,16 +3182,21 @@ public int minDistance(String word1, String word2) {
int[][] dp = new int[m + 1][n + 1];
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0 || j == 0) continue;
dp[i][j] = word1.charAt(i - 1) == word2.charAt(j - 1) ?
dp[i - 1][j - 1] + 1 : Math.max(dp[i][j - 1], dp[i - 1][j]);
if (i == 0 || j == 0) {
continue;
}
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
}
}
}
return m + n - 2 * dp[m][n];
}
```
**修改一个字符串成为另一个字符串**
**编辑距离**
[72. Edit Distance (Hard)](https://leetcode.com/problems/edit-distance/description/)
@ -3194,7 +3205,7 @@ Example 1:
Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')
@ -3202,7 +3213,7 @@ Example 2:
Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
@ -3210,6 +3221,8 @@ exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')
```
题目描述：修改一个字符串成为另一个字符串，使得修改次数最少。一次修改操作包括：插入一个字符、删除一个字符、替换一个字符。
```java
public int minDistance(String word1, String word2) {
if (word1 == null || word2 == null) {
@ -3248,8 +3261,8 @@ public int minDistance(String word1, String word2) {
public int integerBreak(int n) {
int[] dp = new int[n + 1];
dp[1] = 1;
for(int i = 2; i <= n; i++) {
for(int j = 1; j <= i - 1; j++) {
for (int i = 2; i <= n; i++) {
for (int j = 1; j <= i - 1; j++) {
dp[i] = Math.max(dp[i], Math.max(j * dp[i - j], j * (i - j)));
}
}
@ -3270,7 +3283,9 @@ public int numSquares(int n) {
for (int i = 1; i <= n; i++) {
int min = Integer.MAX_VALUE;
for (int square : squareList) {
if (square > i) break;
if (square > i) {
break;
}
min = Math.min(min, dp[i - square] + 1);
}
dp[i] = min;
@ -3299,17 +3314,25 @@ private List<Integer> generateSquareList(int n) {
```java
public int numDecodings(String s) {
if(s == null || s.length() == 0) return 0;
if (s == null || s.length() == 0) {
return 0;
}
int n = s.length();
int[] dp = new int[n + 1];
dp[0] = 1;
dp[1] = s.charAt(0) == '0' ? 0 : 1;
for(int i = 2; i <= n; i++) {
for (int i = 2; i <= n; i++) {
int one = Integer.valueOf(s.substring(i - 1, i));
if(one != 0) dp[i] += dp[i - 1];
if(s.charAt(i - 2) == '0') continue;
if (one != 0) {
dp[i] += dp[i - 1];
}
if (s.charAt(i - 2) == '0') {
continue;
}
int two = Integer.valueOf(s.substring(i - 2, i));
if(two <= 26) dp[i] += dp[i - 2];
if (two <= 26) {
dp[i] += dp[i - 2];
}
}
return dp[n];
}
@ -3342,8 +3365,8 @@ public int uniquePaths(int m, int n) {
```java
public int uniquePaths(int m, int n) {
int S = m + n - 2; // 总共的移动次数
int D = m - 1; // 向下的移动次数
int S = m + n - 2; // 总共的移动次数
int D = m - 1; // 向下的移动次数
long ret = 1;
for (int i = 1; i <= D; i++) {
ret = ret * (S - D + i) / i;
@ -3367,14 +3390,20 @@ Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes
```java
public int minPathSum(int[][] grid) {
if (grid.length == 0 || grid[0].length == 0) return 0;
if (grid.length == 0 || grid[0].length == 0) {
return 0;
}
int m = grid.length, n = grid[0].length;
int[] dp = new int[n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (j == 0) dp[0] = dp[0] + grid[i][0]; // 只能从上侧走到该位置
else if (i == 0) dp[j] = dp[j - 1] + grid[0][j]; // 只能从右侧走到该位置
else dp[j] = Math.min(dp[j - 1], dp[j]) + grid[i][j];
if (j == 0) {
dp[0] = dp[0] + grid[i][0]; // 只能从上侧走到该位置
} else if (i == 0) {
dp[j] = dp[j - 1] + grid[0][j]; // 只能从右侧走到该位置
} else {
dp[j] = Math.min(dp[j - 1], dp[j]) + grid[i][j];
}
}
}
return dp[n - 1];
@ -3517,11 +3546,11 @@ public int minSteps(int n) {
### 素数
（一）素数分解
**素数分解**
每一个数都可以分解成素数的乘积，例如 84 = 2<sup>2</sup> \* 3<sup>1</sup> \* 5<sup>0</sup> \* 7<sup>1</sup> \* 11<sup>0</sup> \* 13<sup>0</sup> \* 17<sup>0</sup> \* …
（二）整除
**整除**
令 x = 2<sup>m0</sup> \* 3<sup>m1</sup> \* 5<sup>m2</sup> \* 7<sup>m3</sup> \* 11<sup>m4</sup> \* …
@ -3529,7 +3558,7 @@ public int minSteps(int n) {
如果 x 整除 y（y mod x == 0），则对于所有 i，mi <= ni。
（三）最大公约数最小公倍数
**最大公约数最小公倍数**
x 和 y 的最大公约数为：gcd(x,y) = 2<sup>min(m0,n0)</sup> \* 3<sup>min(m1,n1)</sup> \* 5<sup>min(m2,n2)</sup> \* ...