45. Jump Game II（跳跃游戏2）

45. Jump Game II（跳跃游戏2）

题目链接

https://leetcode.com/problems/jump-game-ii/description/

题目描述

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.

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.

题目分析

这道题目看起来很简单，可是————
这道题的评级是hard，通过率只有26.2%！！！
我就隐隐觉得这道题绝不是表面看上去那么简单的。
这道题主要是测例的数据量很大算法复杂度不能太高，不然过不了。
直接BFS方法没错，但是最后一组数据会超时。这道题的标签是贪心，所以用贪心就对了。

方法：贪心

算法描述

1. 从起点开始，扫描它能跳到的所有点，如果跳到这个点能延伸它的跳跃范围，那么就跳到这一点。
2. 重复步骤1，直到能跳到终点位置

参考代码

class Solution {
public:
    int jump(vector<int>& nums) {
        int level = 0, cur = 0, next, maxLength = nums[0];
        while (maxLength < nums.size() - 1) {
            level++;
            int max = maxLength;
            for (int i = cur + 1; i <= maxLength && i < nums.size(); i++)
                if (nums[i] + i > max) {
                    max = nums[i] + i;
                    next = i;
                }
            cur = next;
            maxLength = nums[next] + next;
        }
        if (cur != nums.size() - 1)
            level++;
        return level;
    }
};