jump game II

来源：互联网发布：单片机推挽输出编辑：程序博客网时间：2024/04/28 12:50

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.

Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]

The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

这道题本来的做法是DP，用dp[i] 表示跳到第i步需要的最小步数，每次检查第i步的时候更新所能达到的范围里面所有的dp[i+k], 1<=k<=A[i]

public class Solution {    public int jump(int[] A) {        // DP solution: O(n^2) in time        int len = A.length;        int[] dp = new int[len];                // dp[i]表示到i需要的最小步数        for(int i=0; i<len; i++){            //更新从下一个到i+A[i]            for(int j=i+1; j<len && j<i+A[i]; j++){                dp[j] = Math.min(dp[i]+1, dp[j]);                if(j+A[j]>=len-1)   return dp[j]+1;              }                }                return dp[len-1];    }}

这种解法的时间最差是O(n^2), 考虑一个长为n的数列，每个元素是n/2, 每次check的时候都需要更新n/2个dp, 一共就是n^2的量级, 最近的LEETCODE中已经被判为超时了，赞一下LEETCODE

底下是一种更好的解法

public class Solution {    public int jump(int[] A) {        int len = A.length;        if(len <= 1)        return 0;                int step = A[0], jump = 1, range = A[0];                for(int i=1; i<len; i++){            range = Math.max(range, i+A[i]);            step --;                        if(step == 0 && i!=len-1){ // 注意这里的i==len-1这个corner case, 很容易忽略！                jump ++;                step = range - i;            }        }                return jump;    }}

时间复杂度是O(n)

0 0