354. Russian Doll Envelopes

You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.

What is the maximum number of envelopes can you Russian doll? (put one inside other)

Example:
Given envelopes = [[5,4],[6,4],[6,7],[2,3]], the maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).

使用DP去做,但是这种问题与一般的接龙型动态规划的题目最主要的一个区别是需要首先对数据进行排序

java

class Solution {
    public int maxEnvelopes(int[][] envelopes) {
        if (envelopes == null || envelopes.length == 0) {
            return 0;
        }
        if (envelopes[0] == null || envelopes[0].length == 0) {
            return 0;
        }
        Comparator<int[]> cmp = new Comparator<int[]>() {
            public int compare(int[] a1, int[] a2) {
                if (a1[0] == a2[0]) {
                    return a1[1] - a2[1];
                } else {
                    return a1[0] - a2[0];
                }
            }  
        };
        Arrays.sort(envelopes, cmp);
        int[] f = new int[envelopes.length];
        for (int i = 0; i < f.length; i++) {
            f[i] = 1;
        }
        for (int i = 1; i < f.length; i++) {
            for (int j = 0; j < i; j++) {
                if (envelopes[i][0] > envelopes[j][0] && 
                    envelopes[i][1] > envelopes[j][1]) {
                    f[i] = Math.max(f[i], f[j] + 1);
                }
            }
        }
        int sum = 0;
        for (int i : f) {
            sum = Math.max(sum, i);
        }
        return sum;
    }
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包

打赏作者

ncst

你的鼓励将是我创作的最大动力

¥1 ¥2 ¥4 ¥6 ¥10 ¥20
扫码支付:¥1
获取中
扫码支付

您的余额不足,请更换扫码支付或充值

打赏作者

实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值