lintcode: 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.

Example

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

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

思路：

max[i]max[i] 表示经过了（或者跳过了）索引i处能达到的最远位置。

如果max[i−1]<imax[i-1] 说明到不了i，直接返回false；

否则，更新 max[i]max[i]，max[i]=max[i−1]>(i+A[i])?max[i−1]:(i+A[i])max[i]=max[i-1]>(i+A[i])?max[i-1]:(i+A[i])。如果max[i]max[i]大于len-1的话如果已经可以到达终点，返回true。

i遍历到len-2即可

其实没有必要用到max数组，只用一个max变量就可以。

class Solution {
public:
/**
* @param A: A list of integers
* @return: The boolean answer
*/
bool canJump(vector<int> A) {
// write you code here
int len=A.size();
int max;
max[0]=A[0];
if(max[0]>=len-1){
return true;
}
for(int i=1;i<len-1;i++){
if(max[i-1]<i){
return false;
}else{
max[i]=max[i-1]>(i+A[i])?max[i-1]:(i+A[i]);
}

if(max[i]>=len-1){
return true;
}
}

}
};