【LeetCode】Jump Game II 解题报告

标签：jump game   greedy   贪心   

【题目】

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.)

题意：给定一个数组，每个元素代表从该位置可以跳过的距离，问从第一个位置跳到最后一个位置至少需要跳多少次。

说明：上一题 Jump Game 要我们求能不能到达最有一个位置，这道题应该隐含一定可以到达最后一个位置，所以不用考虑不能到达最后一个位置的情况。

思路：依旧是贪心，只不过需要多出一个数组来记录跳的路径。

在 【LeetCode】Jump Game 解题报告 基础上，改进代码如下：

public class Solution {
    public int jump(int[] A) {
        int[] pre = new int[A.length]; //pre[i] keep the last postion which can jump to i
        int reach = 0;
        for (int i = 0; i < A.length; i++) {
            if (i + A[i] > reach) {
                // reach+1 ... i+A[i] are reached by jumping from i
                for (int j = reach + 1; j <= i + A[i] && j < A.length; j++) {
                    pre[j] = i;
                }
                reach = i + A[i];
            }
        }
        
        int ans = 0;
        int k = A.length - 1;
        while (k > 0) {
            k = pre[k];
            ans++;
        }
        return ans;
    }
}

【LeetCode】Jump Game II 解题报告

标签：jump game   greedy   贪心   

原文地址：http://blog.csdn.net/ljiabin/article/details/41826213