[数据结构与算法]【leetcode_942】【简单】di-string-match / 增减字符串匹配

URL

链接：https://leetcode.cn/problems/di-string-match/

题目

分析

源码

工程结构

.
└── main_942.c

源码文件

/* main_942.c */
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int * di_string_match(char * s, int * return_size){
    int n = strlen(s);
    int lo = 0, hi = n;
    int *perm = (int *)malloc(sizeof(int) * (n+1));
    for (int i = 0; i < n; ++i){
        perm[i] = (s[i] == 'I' ? lo++ : hi--);
    }
    perm[n] = lo;// lo == hi
    *return_size = n + 1;
    return perm;
}


int main()
{
    char data[][8]={
        "IDID", // -> [0,4,1,3,2]
        "III",  // -> [0,1,2,3]
        "DDI"   // -> [3,2,0,1]
    };
    int size = sizeof(data)/sizeof(data[0]);
    for(int i = 0; i < size; ++i){
        int ret_size = -1;
        int * p = NULL;
        p = di_string_match(data[i],&ret_size);
        printf("data[%d] = \"%s\", perm = [ ",i,data[i]);
        for (int j = 0; j < ret_size; j++){
            printf("%d ", p[j]);
        }
        printf("]\n");
        if(p != NULL){
            free(p);
        }
    }

    return 0;
}

LOG参考

data[0] = "IDID", perm = [ 0 4 1 3 2 ]
data[1] = "III", perm = [ 0 1 2 3 ]
data[2] = "DDI", perm = [ 3 2 0 1 ]

源码概述

略

小结

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