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Commit 697b8ac

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Merge branch 'master' of github.com:youngyangyang04/leetcode-master
2 parents ca4f55a + 7122ee5 commit 697b8ac

24 files changed

+844
-318
lines changed

‎problems/0005.最长回文子串.md‎

Lines changed: 58 additions & 13 deletions
Original file line numberDiff line numberDiff line change
@@ -263,21 +263,38 @@ public:
263263
Java:
264264

265265
```java
266-
public int[] twoSum(int[] nums, int target) {
267-
int[] res = new int[2];
268-
if(nums == null || nums.length == 0){
269-
return res;
270-
}
271-
Map<Integer, Integer> map = new HashMap<>();
272-
for(int i = 0; i < nums.length; i++){
273-
int temp = target - nums[i];
274-
if(map.containsKey(temp)){
275-
res[1] = i;
276-
res[0] = map.get(temp);
266+
// 双指针 动态规划
267+
class Solution {
268+
public String longestPalindrome(String s) {
269+
if (s.length() == 0 || s.length() == 1) return s;
270+
int length = 1;
271+
int index = 0;
272+
boolean[][] palindrome = new boolean[s.length()][s.length()];
273+
for (int i = 0; i < s.length(); i++) {
274+
palindrome[i][i] = true;
275+
}
276+
277+
for (int L = 2; L <= s.length(); L++) {
278+
for (int i = 0; i < s.length(); i++) {
279+
int j = i + L - 1;
280+
if (j >= s.length()) break;
281+
if (s.charAt(i) != s.charAt(j)) {
282+
palindrome[i][j] = false;
283+
} else {
284+
if (j - i < 3) {
285+
palindrome[i][j] = true;
286+
} else {
287+
palindrome[i][j] = palindrome[i + 1][j - 1];
288+
}
289+
}
290+
if (palindrome[i][j] && j - i + 1 > length) {
291+
length = j - i + 1;
292+
index = i;
293+
}
294+
}
277295
}
278-
map.put(nums[i], i);
296+
return s.substring(index, index + length);
279297
}
280-
return res;
281298
}
282299
```
283300

@@ -363,6 +380,34 @@ class Solution:
363380
Go:
364381

365382
```go
383+
func longestPalindrome(s string) string {
384+
maxLen := 0
385+
left := 0
386+
length := 0
387+
dp := make([][]bool, len(s))
388+
for i := 0; i < len(s); i++ {
389+
dp[i] = make([]bool,len(s))
390+
}
391+
for i := len(s)-1; i >= 0; i-- {
392+
for j := i; j < len(s); j++ {
393+
if s[i] == s[j]{
394+
if j-i <= 1{ // 情况一和情况二
395+
length = j-i
396+
dp[i][j]=true
397+
}else if dp[i+1][j-1]{ // 情况三
398+
length = j-i
399+
dp[i][j] = true
400+
}
401+
}
402+
}
403+
if length > maxLen {
404+
maxLen = length
405+
left = i
406+
}
407+
}
408+
return s[left: left+maxLen+1]
409+
}
410+
366411

367412
```
368413

‎problems/0027.移除元素.md‎

Lines changed: 21 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -146,18 +146,14 @@ public:
146146
};
147147
```
148148

149-
149+
150150
## 相关题目推荐
151151

152152
* 26.删除排序数组中的重复项
153153
* 283.移动零
154154
* 844.比较含退格的字符串
155155
* 977.有序数组的平方
156156

157-
158-
159-
160-
161157
## 其他语言版本
162158

163159

@@ -177,6 +173,26 @@ class Solution {
177173
}
178174
}
179175
```
176+
```java
177+
//相向双指针法
178+
class Solution {
179+
public int removeElement(int[] nums, int val) {
180+
int left = 0;
181+
int right = nums.length - 1;
182+
while(right >= 0 && nums[right] == val) right--; //将right移到从右数第一个值不为val的位置
183+
while(left <= right) {
184+
if(nums[left] == val) { //left位置的元素需要移除
185+
//将right位置的元素移到left(覆盖),right位置移除
186+
nums[left] = nums[right];
187+
right--;
188+
}
189+
left++;
190+
while(right >= 0 && nums[right] == val) right--;
191+
}
192+
return left;
193+
}
194+
}
195+
```
180196

181197
Python:
182198

‎problems/0035.搜索插入位置.md‎

Lines changed: 25 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -226,7 +226,32 @@ class Solution {
226226
}
227227
}
228228
```
229+
```java
230+
//第二种二分法:左闭右开
231+
public int searchInsert(int[] nums, int target) {
232+
int left = 0;
233+
int right = nums.length;
234+
while (left < right) { //左闭右开 [left, right)
235+
int middle = left + ((right - left) >> 1);
236+
if (nums[middle] > target) {
237+
right = middle; // target 在左区间,在[left, middle)中
238+
} else if (nums[middle] < target) {
239+
left = middle + 1; // target 在右区间,在 [middle+1, right)中
240+
} else { // nums[middle] == target
241+
return middle; // 数组中找到目标值的情况,直接返回下标
242+
}
243+
}
244+
// 目标值在数组所有元素之前 [0,0)
245+
// 目标值插入数组中的位置 [left, right) ,return right 即可
246+
// 目标值在数组所有元素之后的情况 [left, right),因为是右开区间,所以 return right
247+
return right;
248+
}
249+
```
250+
251+
252+
229253
Golang:
254+
230255
```golang
231256
// 第一种二分法
232257
func searchInsert(nums []int, target int) int {

‎problems/0051.N皇后.md‎

Lines changed: 51 additions & 50 deletions
Original file line numberDiff line numberDiff line change
@@ -222,56 +222,6 @@ public:
222222

223223
## 其他语言补充
224224

225-
226-
### Python
227-
228-
```python
229-
class Solution:
230-
def solveNQueens(self, n: int) -> List[List[str]]:
231-
if not n: return []
232-
board = [['.'] * n for _ in range(n)]
233-
res = []
234-
def isVaild(board,row, col):
235-
#判断同一列是否冲突
236-
for i in range(len(board)):
237-
if board[i][col] == 'Q':
238-
return False
239-
# 判断左上角是否冲突
240-
i = row -1
241-
j = col -1
242-
while i>=0 and j>=0:
243-
if board[i][j] == 'Q':
244-
return False
245-
i -= 1
246-
j -= 1
247-
# 判断右上角是否冲突
248-
i = row - 1
249-
j = col + 1
250-
while i>=0 and j < len(board):
251-
if board[i][j] == 'Q':
252-
return False
253-
i -= 1
254-
j += 1
255-
return True
256-
257-
def backtracking(board, row, n):
258-
# 如果走到最后一行,说明已经找到一个解
259-
if row == n:
260-
temp_res = []
261-
for temp in board:
262-
temp_str = "".join(temp)
263-
temp_res.append(temp_str)
264-
res.append(temp_res)
265-
for col in range(n):
266-
if not isVaild(board, row, col):
267-
continue
268-
board[row][col] = 'Q'
269-
backtracking(board, row+1, n)
270-
board[row][col] = '.'
271-
backtracking(board, 0, n)
272-
return res
273-
```
274-
275225
### Java
276226

277227
```java
@@ -341,6 +291,55 @@ class Solution {
341291
}
342292
```
343293

294+
### Python
295+
296+
```python
297+
class Solution:
298+
def solveNQueens(self, n: int) -> List[List[str]]:
299+
if not n: return []
300+
board = [['.'] * n for _ in range(n)]
301+
res = []
302+
def isVaild(board,row, col):
303+
#判断同一列是否冲突
304+
for i in range(len(board)):
305+
if board[i][col] == 'Q':
306+
return False
307+
# 判断左上角是否冲突
308+
i = row -1
309+
j = col -1
310+
while i>=0 and j>=0:
311+
if board[i][j] == 'Q':
312+
return False
313+
i -= 1
314+
j -= 1
315+
# 判断右上角是否冲突
316+
i = row - 1
317+
j = col + 1
318+
while i>=0 and j < len(board):
319+
if board[i][j] == 'Q':
320+
return False
321+
i -= 1
322+
j += 1
323+
return True
324+
325+
def backtracking(board, row, n):
326+
# 如果走到最后一行,说明已经找到一个解
327+
if row == n:
328+
temp_res = []
329+
for temp in board:
330+
temp_str = "".join(temp)
331+
temp_res.append(temp_str)
332+
res.append(temp_res)
333+
for col in range(n):
334+
if not isVaild(board, row, col):
335+
continue
336+
board[row][col] = 'Q'
337+
backtracking(board, row+1, n)
338+
board[row][col] = '.'
339+
backtracking(board, 0, n)
340+
return res
341+
```
342+
344343

345344
### Go
346345
```Go
@@ -396,6 +395,8 @@ func isValid(n, row, col int, chessboard [][]string) bool {
396395
return true
397396
}
398397
```
398+
399+
399400
### Javascript
400401
```Javascript
401402
var solveNQueens = function(n) {

‎problems/0053.最大子序和(动态规划).md‎

Lines changed: 14 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -120,6 +120,20 @@ Java:
120120
return res;
121121
}
122122
```
123+
```Java
124+
//因为dp[i]的递推公式只与前一个值有关,所以可以用一个变量代替dp数组,空间复杂度为O(1)
125+
class Solution {
126+
public int maxSubArray(int[] nums) {
127+
int res = nums[0];
128+
int pre = nums[0];
129+
for(int i = 1; i < nums.length; i++) {
130+
pre = Math.max(pre + nums[i], nums[i]);
131+
res = Math.max(res, pre);
132+
}
133+
return res;
134+
}
135+
}
136+
```
123137

124138
Python:
125139
```python

‎problems/0054.螺旋矩阵.md‎

Lines changed: 68 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -133,5 +133,73 @@ public:
133133

134134
## 其他语言版本
135135

136+
### Java
137+
138+
```java
139+
class Solution {
140+
public List<Integer> spiralOrder(int[][] matrix) {
141+
//存放数组的数
142+
List<Integer> ans = new ArrayList<>();
143+
//列数
144+
int columns = matrix[0].length;
145+
//行数
146+
int rows = matrix.length;
147+
//遍历起点
148+
int start = 0;
149+
//循环的次数 行数和列数中的最小值除以二
150+
int loop = Math.min(rows,columns) / 2;
151+
//未遍历的中间列(行)的列(行)下标
152+
int mid = loop;
153+
//终止条件
154+
int offSet = 1;
155+
int i,j;
156+
while(loop-- > 0) {
157+
//初始化起点
158+
i = j = start;
159+
160+
//从左往右
161+
for(; j < columns - offSet; j++)
162+
ans.add(matrix[i][j]);
163+
164+
//从上往下
165+
for(; i < rows - offSet; i++)
166+
ans.add(matrix[i][j]);
167+
168+
//从右往左
169+
for(; j > start; j--)
170+
ans.add(matrix[i][j]);
171+
172+
//从下往上
173+
for(; i > start; i--)
174+
ans.add(matrix[i][j]);
175+
176+
//每循环一次 改变起点位置
177+
start++;
178+
//终止条件改变
179+
offSet++;
180+
}
181+
182+
//如果行和列中的最小值是奇数 则会产生中间行或者中间列没有遍历
183+
if(Math.min(rows,columns) % 2 != 0) {
184+
//行大于列则产生中间列
185+
if(rows > columns) {
186+
//中间列的行的最大下标的下一位的下标为mid + rows - columns + 1
187+
for(int k = mid; k < mid + rows - columns + 1; k++) {
188+
ans.add(matrix[k][mid]);
189+
}
190+
}else {//列大于等于行则产生中间行
191+
//中间行的列的最大下标的下一位的下标为mid + columns - rows + 1
192+
for(int k = mid; k < mid + columns - rows + 1; k++) {
193+
ans.add(matrix[mid][k]);
194+
}
195+
}
196+
}
197+
return ans;
198+
}
199+
}
200+
```
201+
202+
203+
136204
-----------------------
137205
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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