leetcode 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.)

思路：

本题可以用贪婪算法来求解，即每次找到下一次可以到达的最远距离。

1）取第一个对应值作为可以到达的最远处maxReach和当前可以到达的最远处curMaxReach

2）然后从当前index到index+maxReach循环，更新curMaxReach

3）循环结束时，step+1，更新maxReach

4）从当前index到index+maxReach，重复上述过程。

时间复杂度：O(n)

实现如下：

class Solution{
public:
	int jump(vector &nums) {
		int size=nums.size();
		if(size<2) return 0;
		int maxReach = nums[0];
		int curMaxReach = nums[0];
		int curStep = 1;
		for(int i = 1; i <= min(size-1, maxReach); i++){
			curMaxReach = max(curMaxReach, i + nums[i]);
			if(i == size - 1){
				return curStep;
			}
			if(i == maxReach){
				maxReach = curMaxReach;
				curStep++;
			}
		}
		return 0;
	}
};