add 053

Author: andavid

README.md

@@ -75,6 +75,7 @@
 | :--: | :------------------------------------------ |
 | 70   | [Climbing Stairs][070]                      |
 | 121  | [Best Time to Buy and Sell Stock][121]      |
+| 53   | [Maximum Subarray][053]                     |
 
 
 [leetcode]: https://leetcode.com/problemset/all/
@@ -116,3 +117,4 @@
 [278]: https://github.com/andavid/leetcode-java/blob/master/note/278/README.md
 [070]: https://github.com/andavid/leetcode-java/blob/master/note/070/README.md
 [121]: https://github.com/andavid/leetcode-java/blob/master/note/121/README.md
+[053]: https://github.com/andavid/leetcode-java/blob/master/note/053/README.md

note/053/README.md

@@ -0,0 +1,40 @@
+# [Maximum Subarray][title]
+
+## Description
+
+Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
+
+For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
+the contiguous subarray [4,-1,2,1] has the largest sum = 6.
+
+**More practice:**
+
+If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
+
+## 思路
+
+为了求整个字符串最大的子序列和，可以先求子问题，得到局部最大值，然后更新全局最大值。
+可以从前往后扫描，假设 dp[i] 表示前 i 个元素的最大子序列和，如果 dp[i] < 0，那么 dp[i+1] 等于第 i+1 个元素的值，如果 dp[i] > 0，那么 dp[i+1] 等于 dp[i] 加上第 i+1 个元素的值，由此得到前 i+1 个元素的局部最大子序列和。和全局最大值进行比较，如果更大则进行更新。
+
+## [完整代码][src]
+
+```java
+class Solution {
+  public int maxSubArray(int[] nums) {
+    if (nums == null || nums.length == 0) return 0;
+
+    int sum = nums[0];
+    int max = nums[0];
+
+    for (int i = 1; i < nums.length; i++) {
+      sum = (sum < 0) ? nums[i] : (sum + nums[i]);
+      if (sum > max) max = sum;
+    }
+
+    return max;
+  }
+}
+```
+
+[title]: https://leetcode.com/problems/maximum-subarray
+[src]: https://github.com/andavid/leetcode-java/blob/master/src/com/andavid/leetcode/_053/Solution.java

src/com/andavid/leetcode/_053/Solution.java

@@ -0,0 +1,21 @@
+class Solution {
+  public int maxSubArray(int[] nums) {
+    if (nums == null || nums.length == 0) return 0;
+
+    int sum = nums[0];
+    int max = nums[0];
+
+    for (int i = 1; i < nums.length; i++) {
+      sum = (sum < 0) ? nums[i] : (sum + nums[i]);
+      if (sum > max) max = sum;
+    }
+
+    return max;
+  }
+
+  public static void main(String[] args) {
+    Solution solution = new Solution();
+    int[] nums = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
+    System.out.println(solution.maxSubArray(nums));
+  }
+}