README.md

942. 增减字符串匹配

English Version

题目描述

由范围 [0,n] 内所有整数组成的 n + 1 个整数的排列序列可以表示为长度为 n 的字符串 s ，其中:

• 如果 perm[i] < perm[i + 1] ，那么 s[i] == 'I' 
• 如果 perm[i] > perm[i + 1] ，那么 s[i] == 'D' 

给定一个字符串 s ，重构排列 perm 并返回它。如果有多个有效排列perm，则返回其中 任何一个 。

 

示例 1：

输入：s = "IDID"
输出：[0,4,1,3,2]

示例 2：

输入：s = "III"
输出：[0,1,2,3]

示例 3：

输入：s = "DDI"
输出：[3,2,0,1]

 

提示：

• 1 <= s.length <= 105
• s 只包含字符 "I" 或 "D"

解法

类似贪心思想，如果当前字母是 I，我们只需要选择当前可选的最小数字，就能保证后面的数字无论怎么排列，当前数字和下一个数字一定是递增关系。D 同理，选择当前可选的最大数字即可

Python3

class Solution:
    def diStringMatch(self, s: str) -> List[int]:
        n = len(s)
        low, high = 0, n
        ans = []
        for i in range(n):
            if s[i] == 'I':
                ans.append(low)
                low += 1
            else:
                ans.append(high)
                high -= 1
        ans.append(low)
        return ans

Java

class Solution {
    public int[] diStringMatch(String s) {
        int n = s.length();
        int low = 0, high = n;
        int[] ans = new int[n + 1];
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == 'I') {
                ans[i] = low++;
            } else {
                ans[i] = high--;
            }
        }
        ans[n] = low;
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> diStringMatch(string s) {
        int n = s.size();
        int low = 0, high = n;
        vector<int> ans(n + 1);
        for (int i = 0; i < n; ++i) {
            if (s[i] == 'I') {
                ans[i] = low++;
            } else {
                ans[i] = high--;
            }
        }
        ans[n] = low;
        return ans;
    }
};

Go

func diStringMatch(s string) []int {
	n := len(s)
	low, high := 0, n
	var ans []int
	for i := 0; i < n; i++ {
		if s[i] == 'I' {
			ans = append(ans, low)
			low++
		} else {
			ans = append(ans, high)
			high--
		}
	}
	ans = append(ans, low)
	return ans
}

TypeScript

function diStringMatch(s: string): number[] {
    const n = s.length;
    const res = new Array(n + 1);
    let low = 0;
    let high = n;
    for (let i = 0; i < n; i++) {
        if (s[i] === 'I') {
            res[i] = low++;
        } else {
            res[i] = high--;
        }
    }
    res[n] = low;
    return res;
}

Rust

impl Solution {
    pub fn di_string_match(s: String) -> Vec<i32> {
        let s = s.as_bytes();
        let n = s.len();
        let mut res = Vec::with_capacity(n + 1);
        let (mut low, mut high) = (-1, (n + 1) as i32);
        for i in 0..n {
            res.push(if s[i] == b'I' {
                low += 1;
                low
            } else {
                high -= 1;
                high
            });
        }
        res.push(low + 1);
        res
    }
}

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