规范格式

leetcode/0053.Maximum-Subarray/53. Maximum Subarray.go

@@ -0,0 +1,48 @@
+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
+}
+
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}

leetcode/0053.Maximum-Subarray/53. Maximum Subarray_test.go

@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question53 struct {
+	para53
+	ans53
+}
+
+// para 是参数
+// one 代表第一个参数
+type para53 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans53 struct {
+	one int
+}
+
+func Test_Problem53(t *testing.T) {
+
+	qs := []question53{
+
+		question53{
+			para53{[]int{-2, 1, -3, 4, -1, 2, 1, -5, 4}},
+			ans53{6},
+		},
+		question53{
+			para53{[]int{2, 7, 9, 3, 1}},
+			ans53{22},
+		},
+
+		question53{
+			para53{[]int{2}},
+			ans53{2},
+		},
+
+		question53{
+			para53{[]int{-1, -2}},
+			ans53{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 53------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans53, q.para53
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxSubArray(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0053.Maximum-Subarray/README.md

@@ -0,0 +1,27 @@
+# [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)`。