给定一个非负整数数组 nums ,你最初位于数组的 第一个下标 。数组中的每个元素代表你在该位置可以跳跃的最大长度。判断你是否能够到达最后一个下标
https://leetcode-cn.com/problems/jump-game/
示例1:
示例2:
提示:
Java解法
package sj.shimmer.algorithm.m2;
/**
* Created by SJ on 2021/2/26.
*/
class D33 {
public static void main(String[] args) {
System.out.println(canJump(new int[]{2, 3, 1, 1, 4}));
System.out.println(canJump(new int[]{3, 2, 1, 0, 4}));
System.out.println(canJump(new int[]{2,0}));
System.out.println(canJump(new int[]{8,2,4,4,4,9,5,2,5,8,8,0,8,6,9,1,1,6,3,5,1,2,6,6,0,4,8,6,0,3,2,8,7,6,5,1,7,0,3,4,8,3,5,9,0,4,0,1,0,5,9,2,0,7,0,2,1,0,8,2,5,1,2,3,9,7,4,7,0,0,1,8,5,6,7,5,1,9,9,3,5,0,7,5}));
}
public static boolean canJump(int[] nums) {
if (nums != null) {
int length = nums.length;
int maxIndex = 0;
for (int i = 0; i < length; i++) {
if (i<=maxIndex) {//可达
maxIndex = Math.max(maxIndex, i + nums[i]);
if (maxIndex>=length-1) {
return true;
}
}else {
return false;
}
}
}
return false;
}
}
官方解
https://leetcode-cn.com/problems/jump-game/solution/tiao-yue-you-xi-by-leetcode-solution/
-
贪心算法
- 时间复杂度:O(n)
- 空间复杂度:O(1)