45-Jump Game II

【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.)

【分析】

1.采用贪心算法

2.每次都在当前能到距离里选择能走最远的（即i+nums[i]最大），下次到i

public class Solution {
    public int jump(int[] nums) {
        int n=nums.length;
        if(n==0||n==1)
            return 0;
        int max=0;
        int jump=0;
        int start=0;
        while(start<n) {
            max=start+nums[start];
            jump++;
            if(max>=n-1)
                return jump;
            int temp=0;
            for(int i=start+1;i<=max;i++) {
                if(i+nums[i]>temp) {
                    temp=i+nums[i];
                    start=i;
                }
            }
        }
        return jump;
    }
}

 

3.重复，直到i+nums[i]>=n-1,即到达

【算法实现】