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
.
public class Solution {
public boolean canJump(int[] nums) {
boolean res = false;
int count = 0;
for (int i = nums.length - 1; i >= 0; i --) {
if (nums[i] >= count) {
res = true;
count = 0;
} else {
res = false;
}
count ++;
}
return res;
}
}
当然也可以从前向后遍历,设一个reach数,从前向后遍历数组,reach取i+nums[i]和reach较大的一个。如果i大于reach了,说明到不了,停止遍历,返回false,如果i等于n,说明可以到,返回true。代码如下:
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;
}