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
.
思路一:贪心法 确定每个点能够达到的最远距离,并在能够达到的最远距离内,继续迭代,计算最远距离,最后判断maxReach 是否到达n-1处。
i <= maxReach 这个判断条件非常关键
class Solution { public: bool canJump(int A[], int n) { int maxReach = 0;// the most rigth we can reach for(int i = 0; i <= maxReach && i < n; i++) { //cout << maxReach << endl; maxReach = max(maxReach, i + A[i]); } return maxReach >= n-1; } };