LeetCode （45） Jump Game II

题目

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

该题目是LeetCode 55 Jump Game的延伸题目， 不仅需要判断能否到达最终地点，而且需要计算出进行的最小跳数。

该题目，调试了几次都没AC，最终参考别人的思想才过，羞愧。。。

ret:目前为止的jump数

curRch:从A[0]进行ret次jump之后达到的最大范围

curMax:从0~i这i+1个A元素中能达到的最大范围

当curRch < i，说明ret次jump已经不足以覆盖当前第i个元素，因此需要增加一次jump，使之达到

记录的curMax。

AC代码

class Solution {
public:
    int jump(vector<int>& nums) {
        int ret = 0;
        int curMax = 0;
        int curRch = 0;
        int n = nums.size();
        for(int i = 0; i < n; i ++)
        {
            if(curRch < i)
            {
                ret ++;
                curRch = curMax;
            }
            curMax = max(curMax, nums[i]+i);
        }
        return ret;
    }
};

GitHub测试程序源码