题解 | #跳台阶#
跳台阶
https://www.nowcoder.com/practice/8c82a5b80378478f9484d87d1c5f12a4
public class Solution {
public int jumpFloor(int target) {
// 存储上一步的步数,target指的索引,索引+1是数组长度
int[] dp = new int[target + 1];
if (target == 0) {
return 0;
} else if (target == 1) {
return 1;
} else if (target == 2) {
return 2;
}
dp[0] = 0;
dp[1] = 1;
dp[2] = 2;
for (int i = 3; i <= target; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[target];
}
}
解题思想:动态规划,由已知求未知,同时dp数组存储已知值,求最终值。
#算法##算法笔记#
查看14道真题和解析