[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.)

采用类似Dp的解法求解，每次对更新边界。

class Solution {public:    int jump(vector<int>& nums) {         vector<int> temp(nums.size(),0);         int right = 0;         for(int i=0; i<nums.size(); ++i){             if(nums[i]+i>right){ //比最右侧大更新解                for(int j=right+1; j<=i+nums[i]&&j<nums.size(); ++j){ //对上一次的右侧边界和新的右侧边界中间更新解                     if(temp[j]==0)                        temp[j] = temp[i]+1;                 }                 right = nums[i]+i;             }         }         return temp[nums.size()-1];    }};