LeetCode-45-Jump Game II

算法描述：

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.

Example:

Input: [2,3,1,1,4]
Output: 2
Explanation: 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.

Note:

You can assume that you can always reach the last index.

解题思路：贪心法。每次达到当前最大能够到达的最大值时，步数加一。

    int jump(vector<int>& nums) {
        if(nums.size()==0) return 0;
        int curMax = 0;
        int res = 0;
        int maxReach= 0;
        for(int i=0; i < nums.size()-1; i++){
            maxReach = max(maxReach, i+nums[i]);
            if(i == curMax){
                res++;
                curMax = maxReach;
            }
        }
        return res;
    }

LeetCode-45-Jump Game II

原文：https://www.cnblogs.com/nobodywang/p/10369688.html