LeetCode 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.

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.

思路一：将所有的情况都考虑进去，基于递归进行暴力搜索

代码如下：

class Solution {public:    void oneJump(vector<int>& nums,int& minJump,int& count,int& pos)    {        if(pos == nums.size()-1)        {            if(minJump > count)            {                minJump = count;                return;            }        }                    for(int i=pos+1;i<=pos+nums[pos];i++)        {            if(i>nums.size()-1)                return;            count++;            oneJump(nums,minJump,count,i);            count--;        }    }        int jump(vector<int>& nums) {        int minJump = 0x7fffffff;        int count = 0;        int pos = 0;        oneJump(nums,minJump,count,pos);        return minJump;        }};结果是用时超时，需要进行剪枝或寻找更好的方法
思路二：仔细分析题目发现，题目关心的是最少通过几步达到，对于具体的到达选择并不关心，并且必然会达到。所以可以将其转换为广度优先算法模型，然后采用贪心原则进行求解，
代码如下：
class Solution {public:    int jump(vector<int>& nums) {        if(nums.size() < 2)            return 0;                    int level=0,currentMax=0,nextMax=0;        int pos=0;        while(currentMax-pos+1>0)        {            level++;            for(;pos<=currentMax;pos++)            {                if(pos+nums[pos] > nextMax)                    nextMax = pos+nums[pos];                if(nextMax >= nums.size()-1)                    return level;            }            currentMax = nextMax;        }        return 0;    }};