Leetcode题集——jump-game

jump-game：

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.

Determine if you are able to reach the last index.

For example:
A =[2,3,1,1,4], return true.

A =[3,2,1,0,4], return false.

解析：设置变量maxi，表示最远能跳多远。将数组从头开始遍历，当i+A[i]大于maxi时，更新maxi。

直到遍历结束，如maxi>=n-1，则返回为真。用到了贪心算法。

 bool canJump(int A[], int n)     {      int maxi=A[0];      for(int i=1;i<n-1;i++)      {//最后一个位置n-1不在循环内       //第i个位置不可到达，返回错误.        if(i>maxi) return false;        if(i+A[i]>maxi)            maxi=i+A[i];      }            if(maxi>=n-1)          return true;       else            return false;    }

jump-gameII:

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. (Jump1step from index 0 to 1, then 3steps to the last index.)

解析：先找到一个能一口气跳到数组末尾的点i，然后查找能一口气跳到点i的点，一直递归，直到返回到原点。

 int jump(int A[], int n)    {     if(n<=1) return 0;     int end=n-1;     int count=0;          while(end)     {        for(int i=0;i<end;i++)        {          if(i+A[i]>=end)          {             count++;             end=i;          }        }     }      return count;    }