75 Sort Colors

# 1. Question

Given an array withnobjects 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?

# 2. Implementation

(1) Partition in Quick Sort
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class Solution {
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public void sortColors(int[] nums) {
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if (nums == null || nums.length == 0) {
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return;
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}
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int redIndex = 0, blueIndex = nums.length - 1;
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for (int i = 0; i <= blueIndex; i++) {
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if (nums[i] == 0) {
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swap(nums, i, redIndex);
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++redIndex;
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}
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else if (nums[i] == 2) {
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swap(nums, i, blueIndex);
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--blueIndex;
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--i;
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}
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}
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}
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public void swap(int[] nums, int i, int j) {
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int temp = nums[i];
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nums[i] = nums[j];
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nums[j] = temp;
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}
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}
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(2) Count Sort
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class Solution {
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public void sortColors(int[] nums) {
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if (nums == null || nums.length == 0) {
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return;
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}
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int n = nums.length;
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int[] count = new int;
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for (int num : nums) {
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count[num]++;
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}
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int index = 0;
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for (int i = 0; i < count.length; i++) {
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while (count[i] > 0) {
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nums[index++] = i;
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--count[i];
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}
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}
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}
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}
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# 3. Time & Space Complexity

Partition in Quick Sort: 时间复杂度O(n), 空间复杂度O(1)
Count Sort: 时间复杂度O(n), 空间复杂度O(n)