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.
解法：
查看当前的maxstep 是否等于0，如果为0，且没有到末尾，则false，
maxstep+curpos>=size-1 true
maxstep--
curpos++;
如果当前curpos所指的step>maxstep,更新maxstep
bool canJump(vector<int>& nums) {
int size=nums.size();
if(size<=1)
return true;
int maxstep=nums[0];
int pos=0;
while(pos<size){
if(maxstep==0)
return false;
maxstep--;
pos++;
if(maxstep+pos>=size-1)
return true;
if(nums[pos]>maxstep)
maxstep=nums[pos];
}
return true;
}
解法二：leetcode解法
bool canJump(int A[], int n) {
int i = 0;
for (int reach = 0; i < n && i <= reach; ++i) reach = max(i + A[i], reach);
return i == n; }
原文：http://searchcoding.blog.51cto.com/1335412/1694900