leetcode讲解--942. DI String Match-优快云博客

本文深入探讨了LeetCode上942题“DIStringMatch”的算法解决方案，详细讲解了如何根据给定的“I”（增加）和“D”（减少）字符串构造一个满足条件的排列数组。通过实例演示，如IDID输入对应[0,4,1,3,2]输出，以及III和DDI等案例，文章提供了易于理解的Java代码实现。

942. DI String Match

Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length.

Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1:

If S[i] == "I", then A[i] < A[i+1]
If S[i] == "D", then A[i] > A[i+1]

Example 1:

Input: "IDID"
Output: [0,4,1,3,2]

Example 2:

Input: "III"
Output: [0,1,2,3]

Example 3:

Input: "DDI"
Output: [3,2,0,1]

Note:

1 <= S.length <= 10000
S only contains characters "I" or "D".

题目地址

算法：默认是递增的，然后根据D修改顺序，每遇到一串D，就reverse一段A。

java代码

class Solution {
    public int[] diStringMatch(String S) {
        int[] A = new int[S.length()+1];
        for(int i=0;i<=S.length();i++){
            A[i] = i;
        }
        char[] c = S.toCharArray();
        int count=0;
        for(int i=0;i<c.length;i++){
            if(c[i]=='I'){
                continue;
            }
            while(i<c.length && c[i]=='D'){
                i++;
                count++;
            }
            reverse(A, i-count, i);
            count=0;
        }
        return A;
    }
    
    private void reverse(int[] A, int begin, int end){
        for(int i=0;i<(end-begin+1)/2;i++){
            int temp=A[begin+i];
            A[begin+i] = A[end-i];
            A[end-i] = temp;
        }
    }
}