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# [LeetCode] 474. Ones and Zeroes #474

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opened this issue May 30, 2019 · 0 comments
Open

# [LeetCode] 474. Ones and Zeroes#474

opened this issue May 30, 2019 · 0 comments

## Comments

### grandyang commented May 30, 2019

 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 `1s` respectively. 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: The given numbers of `0s` and `1s` will both not exceed `100` 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". ``````   这道题是一道典型的应用DP来解的题，如果我们看到这种求总数，而不是列出所有情况的题，十有八九都是用DP来解，重中之重就是在于找出递推式。如果你第一反应没有想到用DP来做，想得是用贪心算法来做，比如先给字符串数组排个序，让长度小的字符串在前面，然后遍历每个字符串，遇到0或者1就将对应的m和n的值减小，这种方法在有的时候是不对的，比如对于{"11", "01", "10"}，m=2，n=2这个例子，我们将遍历完“11”的时候，把1用完了，那么对于后面两个字符串就没法处理了，而其实正确的答案是应该组成后面两个字符串才对。所以我们需要建立一个二维的DP数组，其中dp[i][j]表示有i个0和j个1时能组成的最多字符串的个数，而对于当前遍历到的字符串，我们统计出其中0和1的个数为zeros和ones，然后dp[i - zeros][j - ones]表示当前的i和j减去zeros和ones之前能拼成字符串的个数，那么加上当前的zeros和ones就是当前dp[i][j]可以达到的个数，我们跟其原有数值对比取较大值即可，所以递推式如下： dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1); 有了递推式，我们就可以很容易的写出代码如下：   ``````class Solution { public: int findMaxForm(vector& strs, int m, int n) { vector> dp(m + 1, vector(n + 1, 0)); for (string str : strs) { int zeros = 0, ones = 0; for (char c : str) (c == '0') ? ++zeros : ++ones; for (int i = m; i >= zeros; --i) { for (int j = n; j >= ones; --j) { dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1); } } } return dp[m][n]; } }; ``````   类似题目： Coin Change   参考资料： https://discuss.leetcode.com/topic/71438/c-dp-solution-with-comments https://discuss.leetcode.com/topic/71417/java-iterative-dp-solution-o-mn-space   LeetCode All in One 题目讲解汇总(持续更新中...)
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