规范格式

leetcode/0209.Minimum-Size-Subarray-Sum/209. Minimum Size Subarray Sum.go

@@ -0,0 +1,32 @@
+package leetcode
+
+func minSubArrayLen(s int, nums []int) int {
+	n := len(nums)
+	if n == 0 {
+		return 0
+	}
+	left, right, res, sum := 0, -1, n+1, 0
+	for left < n {
+		if (right+1) < n && sum < s {
+			right++
+			sum += nums[right]
+		} else {
+			sum -= nums[left]
+			left++
+		}
+		if sum >= s {
+			res = min(res, right-left+1)
+		}
+	}
+	if res == n+1 {
+		return 0
+	}
+	return res
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}

leetcode/0209.Minimum-Size-Subarray-Sum/209. Minimum Size Subarray Sum_test.go

@@ -0,0 +1,43 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question209 struct {
+	para209
+	ans209
+}
+
+// para 是参数
+// one 代表第一个参数
+type para209 struct {
+	s   int
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans209 struct {
+	one int
+}
+
+func Test_Problem209(t *testing.T) {
+
+	qs := []question209{
+
+		question209{
+			para209{7, []int{2, 3, 1, 2, 4, 3}},
+			ans209{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 209------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans209, q.para209
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minSubArrayLen(p.s, p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0209.Minimum-Size-Subarray-Sum/README.md

@@ -0,0 +1,26 @@
+# [209. Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum/)
+
+## 题目
+
+Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead.
+
+Example 1:
+
+```c
+Input: s = 7, nums = [2,3,1,2,4,3]
+Output: 2
+Explanation: the subarray [4,3] has the minimal length under the problem constraint.
+```
+
+Follow up:       
+  
+If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n). 
+
+## 题目大意
+
+给定一个整型数组和一个数字 s，找到数组中最短的一个连续子数组，使得连续子数组的数字之和 sum>=s，返回最短的连续子数组的返回值。
+
+## 解题思路
+
+这一题的解题思路是用滑动窗口。在滑动窗口 [i,j]之间不断往后移动，如果总和小于 s，就扩大右边界 j，不断加入右边的值，直到 sum > s，之和再缩小 i 的左边界，不断缩小直到 sum < s，这时候右边界又可以往右移动。以此类推。
+