Leetcode -- Jump Game II

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.)

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

分析：

该解法采用三个变量，其中

res表示需要走的步数；

currange为当前res步之后所能达到的最大的范围；

curmax为在第i个格子处所能到达的最大范围。

那么，当currange<i 时，说明res步已经无法达到第i个格子了，这时候res+1，范围更新为当前的curmax，也就是在i-1处时所能达到的最大范围。在nums的数值非零时，可以保证在i-1处的curmax>=i，从而更新之后的currange>=i.