shuwenhe
diff --git a/‎0001.leetcode-twoSum-两数之和/main.go
Lines changed: 5 additions & 5 deletions b/‎0001.leetcode-twoSum-两数之和/main.go
Lines changed: 5 additions & 5 deletions
diff --git a/‎0002.leetcode-addTwoNumbers-两数相加/main.go
Lines changed: 0 additions & 101 deletions b/‎0002.leetcode-addTwoNumbers-两数相加/main.go
Lines changed: 0 additions & 101 deletions
diff --git a/‎0007.leetcode-reverse-整数反转/main.go
Lines changed: 26 additions & 0 deletions b/‎0007.leetcode-reverse-整数反转/main.go
Lines changed: 26 additions & 0 deletions
diff --git a/‎0832.leetcode-flipAndInvertImage-翻转图像/main.go
Lines changed: 0 additions & 18 deletions b/‎0832.leetcode-flipAndInvertImage-翻转图像/main.go
Lines changed: 0 additions & 18 deletions
diff --git a/‎shuwen-leetcode.xmind
11.9 KB b/‎shuwen-leetcode.xmind
11.9 KB
@@ -18,12 +18,12 @@ func main() {
 }
 
 func twoSum(nums []int, target int) []int { // 哈希查找的时间复杂度为 O(1)
-	map1 := make(map[int]int) // 哈希容器map降低时间复杂度
-	for index, val := range nums {
-		if wanted, ok := map1[val]; ok { // 判断val1是否在map中，第一次map中肯定没有，所以ok为false，将数组中的值val和位置pos对应的键值对放入map中
-			return []int{wanted, index}
+	m := make(map[int]int) // 哈希容器map降低时间复杂度
+	for k, v := range nums {
+		if index, ok := m[v]; ok { // 判断val1是否在map中，第一次map中肯定没有，所以ok为false，将数组中的值val和位置pos对应的键值对放入map中
+			return []int{index, k}
 		}
-		map1[target-val] = index // 通过hash表查找使val2 = target - val1 的值，如果找到则返回结果，找不到，则将当前值插入map
+		m[target-v] = k // 通过hash表查找使val2 = target - val1 的值，如果找到则返回结果，找不到，则将当前值插入map
 	}
 	return nil
 }
@@ -0,0 +1,26 @@
+// 给出一个 32 位的有符号整数，你需要将这个整数中每位上的数字进行反转。
+// 输入: 123
+// 输出: 321
+
+package main
+
+import (
+	"fmt"
+)
+
+func main() {
+	res := reverse(123)
+	fmt.Println("res = ", res)
+}
+
+func reverse(x int) int {
+	tmp := 0
+	for x != 0 {
+		tmp = tmp*10 + x%10
+		x = x / 10
+	}
+	if tmp > 1<<31-1 || tmp < -(1<<31) {
+		return 0
+	}
+	return tmp
+}
Original file line number	Diff line number	Diff line change
`@@ -18,12 +18,12 @@ func main() {`
`18`	`18`	`}`
`19`	`19`
`20`	`20`	`func twoSum(nums []int, target int) []int { // 哈希查找的时间复杂度为 O(1)`
`21`		`- map1 := make(map[int]int) // 哈希容器map降低时间复杂度`
`22`		`- for index, val := range nums {`
`23`		`- if wanted, ok := map1[val]; ok { // 判断val1是否在map中，第一次map中肯定没有，所以ok为false，将数组中的值val和位置pos对应的键值对放入map中`
`24`		`- return []int{wanted, index}`
	`21`	`+ m := make(map[int]int) // 哈希容器map降低时间复杂度`
	`22`	`+ for k, v := range nums {`
	`23`	`+ if index, ok := m[v]; ok { // 判断val1是否在map中，第一次map中肯定没有，所以ok为false，将数组中的值val和位置pos对应的键值对放入map中`
	`24`	`+ return []int{index, k}`
`25`	`25`	`}`
`26`		`- map1[target-val] = index // 通过hash表查找使val2 = target - val1 的值，如果找到则返回结果，找不到，则将当前值插入map`
	`26`	`+ m[target-v] = k // 通过hash表查找使val2 = target - val1 的值，如果找到则返回结果，找不到，则将当前值插入map`
`27`	`27`	`}`
`28`	`28`	`return nil`
`29`	`29`	`}`