[LeetCode] 45. Jump Game II Java

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.)

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

题意及分析：使用贪心算法，每次找到该点前能到达的最大点lastreach，如果lastreach大于i，那么重新计算能达到的最大点，否则step+1，重置lastSearch,重新计算下一步能达到的最大点。

代码：

class Solution {
    public int jump(int[] nums) {
        if (nums.length <= 1) return 0;
        int reach = nums[0];       //记录从当前点能达到的最大位置
        int lastreach = 0;      //记录前面能到达的最大点
        int step = 0;

        for(int i=1;i<=reach && i<nums.length;i++){
            if(i > lastreach){
                step++;
                lastreach = reach;
            }
            reach = Math.max(reach, i+nums[i]);
        }
        if(reach < nums.length-1) return 0;
        return step;
    }
}

 

转载于:https://www.cnblogs.com/271934Liao/p/8334044.html