[LEET-474] add 474

fishercoder1534 · fishercoder1534 · commit 2bbbd5a99116 · 2017-02-14T08:48:21.000-08:00
diff --git a/leetcode-algorithms/README.md b/leetcode-algorithms/README.md
@@ -13,6 +13,7 @@
 |485|[Max Consecutive Ones](https://leetcode.com/problems/max-consecutive-ones/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/MaxConsecutiveOnes.java) | O(n) |O(1) | Easy| Array
 |477|[Total Hamming Distance](https://leetcode.com/problems/total-hamming-distance/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/TotalHammingDistance.java) | O(n) |O(1) | Medium| Bit Manipulation
 |476|[Number Complement](https://leetcode.com/problems/number-complement/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/NumberComplement.java) | O(n) |O(1) | Easy| Bit Manipulation
+|474|[Ones and Zeroes](https://leetcode.com/problems/ones-and-zeroes/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/OnesandZeroes.java) | O(n) |O(m*n) | Medium| DP
 |472|[Concatenated Words](https://leetcode.com/problems/concatenated-words/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/ConcatenatedWords.java) | O(n^2) |O(n) | Hard| Trie, DP, DFS
 |467|[Unique Substrings in Wraparound String](https://leetcode.com/problems/unique-substrings-in-wraparound-string/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/UniqueSubstringsinWraparoundString.java) | O(n) |O(1) | Medium| DP
 |463|[Island Perimeter](https://leetcode.com/problems/island-perimeter/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/IslandPerimeter.java)| O(m*n)|O(1) | Easy| 
diff --git a/leetcode-algorithms/src/main/java/com/stevesun/solutions/OnesandZeroes.java b/leetcode-algorithms/src/main/java/com/stevesun/solutions/OnesandZeroes.java
@@ -0,0 +1,53 @@
+package com.stevesun.solutions;
+
+/**
+ * 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".
+
+ */
+public class OnesandZeroes {
+
+    public int findMaxForm(String[] strs, int m, int n) {
+        int[][] dp = new int[m+1][n+1];
+        for (String str : strs){
+            int[] count = count(str);
+            for (int i = m; i >= count[0]; i--){
+                for (int j = n; j >= count[1]; j--){
+                    dp[i][j] = Math.max(dp[i][j], dp[i - count[0]][j - count[1]] + 1);
+                }
+            }
+        }
+        return dp[m][n];
+    }
+
+    private int[] count(String str) {
+        int[] count = new int[]{};
+        for (char c : str.toCharArray()){
+            count[c - '0']++;
+        }
+        return count;
+    }
+
+}
