SortColors I && II

Change-Id: I333d915b135f867703e379f4c61fe80bbc29c088

SortColors.java

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+/*
+ Author:     King, [email protected]
+ Date:       Dec 24, 2013
+ Problem:    Sort Colors
+ Difficulty: Medium
+ Source:     https://oj.leetcode.com/problems/sort-colors/
+ Notes:
+ Given an array with n objects colored red, white or blue, sort them so that objects of the same color
+ are adjacent, with the colors in the order red, white and blue.
+ Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.
+ Note:
+ You are not suppose to use the library's sort function for this problem.
+ Follow up:
+ A rather straight forward solution is a two-pass algorithm using counting sort.
+ First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with 
+ total number of 0's, then 1's and followed by 2's.
+ Could you come up with an one-pass algorithm using only constant space?
+
+ Solution: 0 0 0 1 1 1 1 ...... 2 2 2 2
+               |         |      |
+             zero        i     two
+              ->        ->     <-  
+ */
+public class Solution {
+    public void sortColors(int[] A) {
+        int n = A.length;
+        if (n <= 1) return;
+        for (int i = 0, left = 0, right = n - 1; i <= right;) {
+            if (A[i] == 0) {
+                A[i++] = A[left];
+                A[left++] = 0;
+            } else if (A[i] == 2) {
+                A[i] = A[right];
+                A[right--] = 2;
+            } else i++;
+        }
+    }
+}

SortColorsII.java

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+/*
+ Author:     King, [email protected]
+ Date:       Dec 24, 2013
+ Problem:    Sort Colors
+ Difficulty: Medium
+ Source:     http://lintcode.com/zh-cn/problem/sort-colors-ii/
+ Notes:
+ Given an array of n objects with k different colors (numbered from 1 to k), sort them so that objects of the same color are adjacent, with the colors in the order 1, 2, ... k.
+
+ Note
+ You are not suppose to use the library's sort function for this problem.
+
+ Example
+ GIven colors=[3, 2, 2, 1, 4], k=4, your code should sort colors in-place to [1, 2, 2, 3, 4]. 
+
+ Challenge
+ A rather straight forward solution is a two-pass algorithm using counting sort. That will cost O(k) extra memory.
+
+ Can you do it without using extra memory?
+
+ Solution: Use the first k buckets to store the count. (count sort, two pass).
+ */
+class Solution {
+    /**
+     * @param colors: A list of integer
+     * @param k: An integer
+     * @return: nothing
+     */
+    public void sortColors2(int[] A, int k) {
+        // write your code here
+        int n = A.length;
+        if (n <= 1) return;
+        for (int i = 0; i < n; ++i) {
+            if(A[i] > 0) {
+                int c = A[i];
+                A[i] = 0;
+                while(true) {
+                    if (A[c-1] <= 0) {
+                        --A[c-1];
+                        break;
+                    } else {
+                        int col = A[c-1];
+                        A[c-1] = -1;
+                        c = col;
+                    }
+                }
+            }
+        }
+        int idx = n;
+        for (int i = k; i > 0; --i) {
+            for (int j = 0; j > A[i-1]; --j) A[--idx] = i;
+        }
+    }
+}