添加内容

website/content/ChapterFour/0053.Maximum-Subarray.md

@@ -0,0 +1,75 @@
+# [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)
+
+
+## 题目
+
+Given an integer array `nums`, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
+
+**Example**:
+
+
+    Input: [-2,1,-3,4,-1,2,1,-5,4],
+    Output: 6
+    Explanation: [4,-1,2,1] has the largest sum = 6.
+
+
+**Follow up**:
+
+If you have figured out the O(*n*) solution, try coding another solution using the divide and conquer approach, which is more subtle.
+
+## 题目大意
+
+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。
+
+## 解题思路
+
+- 这一题可以用 DP 求解也可以不用 DP。
+- 题目要求输出数组中某个区间内数字之和最大的那个值。`dp[i]` 表示 `[0,i]` 区间内各个子区间和的最大值，状态转移方程是 `dp[i] = nums[i] + dp[i-1] (dp[i-1] > 0)`，`dp[i] = nums[i] (dp[i-1] ≤ 0)`。
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一 DP
+func maxSubArray(nums []int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	if len(nums) == 1 {
+		return nums[0]
+	}
+	dp, res := make([]int, len(nums)), nums[0]
+	dp[0] = nums[0]
+	for i := 1; i < len(nums); i++ {
+		if dp[i-1] > 0 {
+			dp[i] = nums[i] + dp[i-1]
+		} else {
+			dp[i] = nums[i]
+		}
+		res = max(res, dp[i])
+	}
+	return res
+}
+
+// 解法二 模拟
+func maxSubArray1(nums []int) int {
+	if len(nums) == 1 {
+		return nums[0]
+	}
+	maxSum, res, p := nums[0], 0, 0
+	for p < len(nums) {
+		res += nums[p]
+		if res > maxSum {
+			maxSum = res
+		}
+		if res < 0 {
+			res = 0
+		}
+		p++
+	}
+	return maxSum
+}
+
+```