LeetCode 474 Ones and Zeroes (二维01背包)

In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.

For now, suppose you are a dominator of m 0s and n 1srespectively. On the other hand, there is an array with strings consisting of only 0s and 1s.

Now your task is to find the maximum number of strings that you can form with given m 0s and n 1s. Each 0 and 1 can be used at most once.

Note:

1. The given numbers of 0s and 1s will both not exceed 100
2. The size of given string array won't exceed 600.

Example 1:

Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3
Output: 4

Explanation: This are totally 4 strings can be formed by the using of 5 0s and 3 1s, which are “10,”0001”,”1”,”0”

Example 2:

Input: Array = {"10", "0", "1"}, m = 1, n = 1
Output: 2

Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".

题目链接：https://leetcode.com/problems/ones-and-zeroes/

题目分析：01背包一眼题，只不过背包有两个维度的限制（0的个数和1的个数），原理相同

以下做法18ms，击败86%

class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int[][] dp = new int[m + 1][n + 1];
        int[] ones = new int[strs.length];
        for (int i = 0; i < strs.length; i++) {
            for (int j = 0; j < strs[i].length(); j++) {
                if (strs[i].charAt(j) == '1') {
                    ones[i]++;
                }
            }
        }
        for (int i = 0; i < strs.length; i++) {
            int zerosi = strs[i].length() - ones[i];
            for (int j = m; j >= zerosi; j--) {
                for (int k = n; k >= ones[i]; k--) {
                    if (dp[j][k] < dp[j - zerosi][k - ones[i]] + 1) {
                        dp[j][k] = dp[j - zerosi][k - ones[i]] + 1;
                    }
                }
            }
        }
        return dp[m][n];
    }
}

其实不难发现，没有必要存ones数组，直接在dp最外层循环计算即可

10ms，击败100%

class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 0; i < strs.length; i++) {
            int onesi = 0;
            for (int j = 0; j < strs[i].length(); j++) {
                if (strs[i].charAt(j) == '1') {
                    onesi++;
                }
            }
            int zerosi = strs[i].length() - onesi;
            for (int j = m; j >= zerosi; j--) {
                for (int k = n; k >= onesi; k--) {
                    if (dp[j][k] < dp[j - zerosi][k - onesi] + 1) {
                        dp[j][k] = dp[j - zerosi][k - onesi] + 1;
                    }
                }
            }
        }
        return dp[m][n];
    }
}