Add solution 053

Author: yanglbme

README.md

@@ -16,6 +16,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 007 | [Reverse Integer](https://github.com/yanglbme/leetcode/tree/master/solution/007.Reverse%20Integer) | `Math` |
 | 013 | [Roman to Integer](https://github.com/yanglbme/leetcode/tree/master/solution/013.Roman%20to%20Integer) | `Math`, `String` |
 | 021 | [Merge Two Sorted Lists](https://github.com/yanglbme/leetcode/tree/master/solution/021.Merge%20Two%20Sorted%20Lists) | `Linked List` |
+| 053 | [Maximum Subarray](https://github.com/yanglbme/leetcode/tree/master/solution/053.Maximum%20Subarray) | `Array`, `Divide and Conquer`, `Dynamic Programming` |
 | 070 | [Climbing Stairs](https://github.com/yanglbme/leetcode/tree/master/solution/070.Climbing%20Stairs) | `Dynamic Programming` |
 | 083 | [Remove Duplicates from Sorted List](https://github.com/yanglbme/leetcode/tree/master/solution/083.Remove%20Duplicates%20from%20Sorted%20List) | `Linked List` |
 | 198 | [House Robber](https://github.com/yanglbme/leetcode/tree/master/solution/198.House%20Robber) | `Dynamic Programming` |

solution/053.Maximum Subarray/README.md

@@ -0,0 +1,80 @@
+## 最大子序和
+### 题目描述
+
+给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。
+
+示例:
+```
+输入: [-2,1,-3,4,-1,2,1,-5,4],
+输出: 6
+解释: 连续子数组 [4,-1,2,1] 的和最大，为 6。
+```
+
+进阶:
+
+如果你已经实现复杂度为 O(n) 的解法，尝试使用更为精妙的分治法求解。
+
+### 解法
+此题可以用动态规划法，开辟一个数组res，res[i] 表示以当前结点nums[i] 结尾的最大连续子数组的和。最后计算 res 的最大元素即可。
+也可以用分治法，最大连续子数组有三种情况：在原数组左侧、右侧、跨中间结点，返回这三者的最大值即可。
+
+
+动态规划法：
+
+```java
+class Solution {
+    public int maxSubArray(int[] nums) {
+        int n = nums.length;
+        if (n == 1) {
+            return nums[0];
+        }
+        int[] res = new int[n];
+        res[0] = nums[0];
+        int max = res[0];
+        for (int i = 1; i < n; ++i) {
+            res[i] = Math.max(res[i - 1] + nums[i], nums[i]);
+            max = Math.max(res[i], max);
+        }
+        
+        return max;
+        
+    }
+}
+```
+
+分治法：
+
+```java
+class Solution {
+    public int maxSubArray(int[] nums) {
+        return maxSubArray(nums, 0, nums.length - 1);
+    }
+    
+    private int maxSubArray(int[] nums, int start, int end) {
+        if (start == end) {
+            return nums[start];
+        }
+        int mid = start + ((end - start) >> 1);
+        int left = maxSubArray(nums, start, mid);
+        int right = maxSubArray(nums, mid + 1, end);
+        
+        int leftSum = 0;
+        int leftMax = Integer.MIN_VALUE;
+        for (int i = mid; i >= start; --i) {
+            leftSum += nums[i];
+            leftMax = Math.max(leftSum, leftMax);
+        }
+        
+        int rightSum = 0;
+        int rightMax = Integer.MIN_VALUE;
+        for (int i = mid + 1; i <= end; ++i) {
+            rightSum += nums[i];
+            rightMax = Math.max(rightSum, rightMax);
+        }
+        
+        return Math.max(Math.max(left, right), leftMax + rightMax);
+        
+        
+    }
+}
+```

solution/053.Maximum Subarray/Solution.java

@@ -0,0 +1,18 @@
+class Solution {
+    public int maxSubArray(int[] nums) {
+        int n = nums.length;
+        if (n == 1) {
+            return nums[0];
+        }
+        int[] res = new int[n];
+        res[0] = nums[0];
+        int max = res[0];
+        for (int i = 1; i < n; ++i) {
+            res[i] = Math.max(res[i - 1] + nums[i], nums[i]);
+            max = Math.max(res[i], max);
+        }
+        
+        return max;
+        
+    }
+}