Add solution 1603、1608、1700、1710、1716、1720、1725、1732、1736、1742、1748、1752、1758

leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/1758. Minimum Changes To Make Alternating Binary String.go

@@ -0,0 +1,18 @@
+package leetcode
+
+func minOperations(s string) int {
+	res := 0
+	for i := 0; i < len(s); i++ {
+		if int(s[i]-'0') != i%2 {
+			res++
+		}
+	}
+	return min(res, len(s)-res)
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}

leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/1758. Minimum Changes To Make Alternating Binary String_test.go

@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1758 struct {
+	para1758
+	ans1758
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1758 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1758 struct {
+	one int
+}
+
+func Test_Problem1758(t *testing.T) {
+
+	qs := []question1758{
+
+		{
+			para1758{"0100"},
+			ans1758{1},
+		},
+
+		{
+			para1758{"10"},
+			ans1758{0},
+		},
+
+		{
+			para1758{"1111"},
+			ans1758{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1758------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1758, q.para1758
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minOperations(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1758.Minimum-Changes-To-Make-Alternating-Binary-String/README.md

@@ -0,0 +1,70 @@
+# [1758. Minimum Changes To Make Alternating Binary String](https://leetcode.com/problems/minimum-changes-to-make-alternating-binary-string/)
+
+
+## 题目
+
+You are given a string `s` consisting only of the characters `'0'` and `'1'`. In one operation, you can change any `'0'` to `'1'` or vice versa.
+
+The string is called alternating if no two adjacent characters are equal. For example, the string `"010"` is alternating, while the string `"0100"` is not.
+
+Return *the **minimum** number of operations needed to make* `s` *alternating*.
+
+**Example 1:**
+
+```
+Input: s = "0100"
+Output: 1
+Explanation: If you change the last character to '1', s will be "0101", which is alternating.
+```
+
+**Example 2:**
+
+```
+Input: s = "10"
+Output: 0
+Explanation: s is already alternating.
+```
+
+**Example 3:**
+
+```
+Input: s = "1111"
+Output: 2
+Explanation: You need two operations to reach "0101" or "1010".
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 104`
+- `s[i]` is either `'0'` or `'1'`.
+
+## 题目大意
+
+你将得到一个仅包含字符“ 0”和“ 1”的字符串 `s`。 在一项操作中，你可以将任何 `'0'` 更改为 `'1'`，反之亦然。 如果两个相邻字符都不相等，则该字符串称为交替字符串。 例如，字符串“ 010”是交替的，而字符串“ 0100”则不是。 返回使 `s` 交替所需的最小操作数。
+
+## 解题思路
+
+- 简单题。利用数组下标奇偶交替性来判断交替字符串。交替字符串有 2 种，一个是 `'01010101……'` 还有一个是 `'1010101010……'`，这两个只需要计算出一个即可，另外一个利用 `len(s) - res` 就是答案。
+
+## 代码
+
+```go
+package leetcode
+
+func minOperations(s string) int {
+	res := 0
+	for i := 0; i < len(s); i++ {
+		if int(s[i]-'0') != i%2 {
+			res++
+		}
+	}
+	return min(res, len(s)-res)
+}
+
+func min(a, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+```