45. 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.

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.

解答：The main idea is based on greedy. Let's say the range of the current jump is [curBegin, curEnd], curFarthest is the farthest point that all points in [curBegin, curEnd] can reach. Once the current point reaches curEnd, then trigger another jump, and set the new curEnd with curFarthest, then keep the above steps, as the following:

public int jump(int[] A) {
    int jumps = 0, curEnd = 0, curFarthest = 0;
    for (int i = 0; i < A.length - 1; i++) {
        curFarthest = Math.max(curFarthest, i + A[i]);
        if (i == curEnd) {
            jumps++;
            curEnd = curFarthest;
        }
    }
    return jumps;
}

 我们的任务是从0到达最后一个点，而不是越过最后一个点，所以i 《 length - 1。curfarthest是当前能到的最远的点，curend是一次跳跃的终点，如果i == curend了说明需要进行下一次跳跃。

class Solution {
    public int jump(int[] nums) {
        int res = 0;
        int curend = 0;
        int curfarest = 0;
        for(int i = 0; i < nums.length; i++){
            curfarest = Math.max(curfarest, i + nums[i]);
            if (curend >= nums.length - 1) {
                break;
            }
            if(i == curend){
                res++;
                curend = curfarest;
            }
        }
        return res;
    }
}

为了工整也可以到最后一个点，但需要判断是否越界