[codility]Frog-River-One

最新推荐文章于 2019-08-17 20:17:15 发布

原创最新推荐文章于 2019-08-17 20:17:15 发布 · 1.5k 阅读

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本文介绍了一种使用C++编程解决青蛙跳过障碍物问题的方法，通过优化数据结构和算法，实现了在给定数组中找到青蛙能够到达的最远位置。利用`algorithm`和`climits`库，通过记录每个位置的最早跌落时间，最终确定青蛙能否到达终点并返回其最短路径。

// you can also use includes, for example:
// #include <algorithm>
#include<climits>
int solution(int X, vector<int> &A) {
    // write your code in C++98
    //...out of range case
    if(X > A.size()) return -1;
    //...keep record of the earlist falling time of each position
    vector<int> timeOfPos(X+1, -1);
    for(int i = 0; i < A.size(); ++i)
        timeOfPos[A[i]] = timeOfPos[A[i]] == -1 ? i : min(i, timeOfPos[A[i]]);
    //...check if the frog can reach to the otherside
    int maxSingleTime = INT_MIN;
    for(int i = 1; i < timeOfPos.size(); ++i)
        if(timeOfPos[i] == -1) return -1;
        else maxSingleTime = max(timeOfPos[i], maxSingleTime);
    //...return the result
    return maxSingleTime;
}