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https://github.com/halfrost/LeetCode-Go.git
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规范格式
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package leetcode
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import (
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"sort"
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"github.com/halfrost/LeetCode-Go/template"
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)
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func smallestStringWithSwaps(s string, pairs [][]int) string {
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uf, res, sMap := template.UnionFind{}, []byte(s), map[int][]byte{}
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uf.Init(len(s))
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for _, pair := range pairs {
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uf.Union(pair[0], pair[1])
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}
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for i := 0; i < len(s); i++ {
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r := uf.Find(i)
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sMap[r] = append(sMap[r], s[i])
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}
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for _, v := range sMap {
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sort.Slice(v, func(i, j int) bool {
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return v[i] < v[j]
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})
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}
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for i := 0; i < len(s); i++ {
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r := uf.Find(i)
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bytes := sMap[r]
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res[i] = bytes[0]
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sMap[r] = bytes[1:len(bytes)]
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}
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return string(res)
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}
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package leetcode
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import (
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"fmt"
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"testing"
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)
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type question1202 struct {
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para1202
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ans1202
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}
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// para 是参数
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// one 代表第一个参数
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type para1202 struct {
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s string
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pairs [][]int
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}
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// ans 是答案
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// one 代表第一个答案
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type ans1202 struct {
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one string
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}
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func Test_Problem1202(t *testing.T) {
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qs := []question1202{
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question1202{
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para1202{"dcab", [][]int{[]int{0, 3}, []int{1, 2}}},
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ans1202{"bacd"},
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},
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question1202{
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para1202{"dcab", [][]int{[]int{0, 3}, []int{1, 2}, []int{0, 2}}},
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ans1202{"abcd"},
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},
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question1202{
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para1202{"cba", [][]int{[]int{0, 1}, []int{1, 2}}},
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ans1202{"abc"},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 1202------------------------\n")
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for _, q := range qs {
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_, p := q.ans1202, q.para1202
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fmt.Printf("【input】:%v 【output】:%v\n", p, smallestStringWithSwaps(p.s, p.pairs))
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}
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fmt.Printf("\n\n\n")
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}
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63
leetcode/1202.Smallest-String-With-Swaps/README.md
Executable file
63
leetcode/1202.Smallest-String-With-Swaps/README.md
Executable file
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# [1202. Smallest String With Swaps](https://leetcode.com/problems/smallest-string-with-swaps/)
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## 题目:
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You are given a string `s`, and an array of pairs of indices in the string `pairs` where `pairs[i] = [a, b]` indicates 2 indices(0-indexed) of the string.
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You can swap the characters at any pair of indices in the given `pairs` **any number of times**.
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Return the lexicographically smallest string that `s` can be changed to after using the swaps.
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**Example 1:**
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Input: s = "dcab", pairs = [[0,3],[1,2]]
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Output: "bacd"
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Explaination:
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Swap s[0] and s[3], s = "bcad"
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Swap s[1] and s[2], s = "bacd"
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**Example 2:**
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Input: s = "dcab", pairs = [[0,3],[1,2],[0,2]]
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Output: "abcd"
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Explaination:
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Swap s[0] and s[3], s = "bcad"
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Swap s[0] and s[2], s = "acbd"
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Swap s[1] and s[2], s = "abcd"
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**Example 3:**
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Input: s = "cba", pairs = [[0,1],[1,2]]
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Output: "abc"
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Explaination:
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Swap s[0] and s[1], s = "bca"
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Swap s[1] and s[2], s = "bac"
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Swap s[0] and s[1], s = "abc"
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**Constraints:**
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- `1 <= s.length <= 10^5`
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- `0 <= pairs.length <= 10^5`
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- `0 <= pairs[i][0], pairs[i][1] < s.length`
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- `s` only contains lower case English letters.
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## 题目大意
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给你一个字符串 s,以及该字符串中的一些「索引对」数组 pairs,其中 pairs[i] = [a, b] 表示字符串中的两个索引(编号从 0 开始)。你可以 任意多次交换 在 pairs 中任意一对索引处的字符。返回在经过若干次交换后,s 可以变成的按字典序最小的字符串。
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提示:
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- 1 <= s.length <= 10^5
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- 0 <= pairs.length <= 10^5
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- 0 <= pairs[i][0], pairs[i][1] < s.length
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- s 中只含有小写英文字母
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## 解题思路
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- 给出一个字符串和一个字符串里可交换的下标。要求交换以后字典序最小的字符串。
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- 这一题可以用并查集来解题,先把可交换下标都 `Union()` 起来,每个集合内,按照字典序从小到大排列。最后扫描原有字符串,从左到右依次找到各自对应的集合里面最小的字符进行替换,每次替换完以后,删除集合中该字符(防止下次重复替换)。最终得到的字符就是最小字典序的字符。
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