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Leetcode:Jump Game II

时间:2015-01-15 12:27:03      阅读:128      评论:0      收藏:0      [点我收藏+]

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Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]

The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

分析:贪心算法。关键在于维护当前step生成的window最右端和step+1生成的window最右端。如果当前window最右端大于等于n-1,那么返回step。代码如下:

class Solution {
public:
    int jump(int A[], int n) {
        int step = 0;
        if(n == 1) return step;
        int right = 0, old_right = 0;
        
        for(int i = 0; i < n; i++){
            if(i > old_right){
                step++;
                old_right = right;
            }
            right = max(right, i+A[i]);
        }
        
        return step;
    }
};

 

Leetcode:Jump Game II

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原文地址:http://www.cnblogs.com/Kai-Xing/p/4225828.html

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