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.
代码
Approach #1: Dynamic Programming
public class Solution {
public boolean canJump(int[] A) {
boolean[] f = new boolean[A.length];
f[0] = true;
for (int i = 1; i < A.length; i++) {
for (int j = 0; j < i; j++) {
if (f[j] && j + A[j] >= i) {
f[i] = true;
break;
}
}
}
return f[A.length - 1];
}
}
Approach #2: Greedy
public class Solution {
public boolean canJump(int[] A) {
if (A == null || A.length == 0) {
return false;
}
int farthest = A[0];
for (int i = 1; i < A.length; i++) {
if (i <= farthest && A[i] + i >= farthest) {
farthest = A[i] + i;
}
}
return farthest >= A.length - 1;
}
}
class Solution {
public boolean canJump(int[] nums) {
int n = nums.length;
int maxPos = nums[0];
for (int i = 1; i < n; i++) {
if (maxPos < i) {
return false;
}
maxPos = Math.max(maxPos, nums[i] + i);
}
return true;
}
}