leetcode-java/src/SortColors75.java at master · fluency03/leetcode-java · GitHub

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/**
 * 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?
 */


importjava.util.Arrays;


publicclassSortColors75 {
publicvoidsortColors(int[] nums) {
intL = nums.length;
if (L == 0 || L == 1) {
return;
 }

helper(0, L - 1, nums);
 }

privatevoidhelper(intleft, intright, int[] nums) {
if (left == right) {
return;
 }

intmid = (left + right) / 2;
helper(left, mid, nums);
helper(mid + 1, right, nums);
merge(left, mid, right, nums);
 }

privatevoidmerge(intleft, intmid, intright, int[] nums) {
intl = 0;
intr = 0;
inti = left;
int[] la = Arrays.copyOfRange(nums, left, mid+1);
int[] ra = Arrays.copyOfRange(nums, mid+1, right+1);
while (l <= mid - left && r <= right - mid - 1) {
if (la[l] <= ra[r]) {
nums[i] = la[l];
l++;
 } else {
nums[i] = ra[r];
r++;
 }
i++;
 }
while (l <= mid - left) {
nums[i] = la[l];
l++;
i++;
 }
 }


/**
 * https://discuss.leetcode.com/topic/6968/four-different-solutions
 */
// two pass O(m+n) space
publicvoidsortColors2(intA[]) {
intnum0 = 0, num1 = 0, num2 = 0;

for(inti = 0; i < A.length; i++) {
if (A[i] == 0) ++num0;
elseif (A[i] == 1) ++num1;
elseif (A[i] == 2) ++num2;
 }

for(inti = 0; i < num0; ++i) A[i] = 0;
for(inti = 0; i < num1; ++i) A[num0+i] = 1;
for(inti = 0; i < num2; ++i) A[num0+num1+i] = 2;
 }

// one pass in place solution
publicvoidsortColors3(intA[]) {
intn0 = -1, n1 = -1, n2 = -1;
for (inti = 0; i < A.length; ++i) {
if (A[i] == 0)
 {
A[++n2] = 2; A[++n1] = 1; A[++n0] = 0;
 }
elseif (A[i] == 1)
 {
A[++n2] = 2; A[++n1] = 1;
 }
elseif (A[i] == 2)
 {
A[++n2] = 2;
 }
 }
 }

// one pass in place solution
publicvoidsortColors4(intA[]) {
intj = 0, k = A.length - 1;
for (inti = 0; i <= k; ++i){
if (A[i] == 0 && i != j)
swap(A, i--, j++);
elseif (A[i] == 2 && i != k)
swap(A, i--, k--);
 }
 }


// one pass in place solution
publicvoidsortColors5(intA[], intn) {
intj = 0, k = n-1;
for (inti=0; i <= k; i++) {
if (A[i] == 0)
swap(A, i, j++);
elseif (A[i] == 2)
swap(A, i--, k--);
 }
 }

// bucket sort
publicvoidsortColors6(int[] nums) {
int[] buckets = newint[3];

for (inti=0; i<nums.length; i++) {
buckets[nums[i]]++;
 }

intidx = 0;
for (inti=0; i<3; i++) {
intcount = buckets[i];
for (intj=0; j<count; j++) {
nums[idx++] = i;
 }
 }

 }

publicvoidsortColors7(int[] nums) {
if (nums == null || nums.length <= 1) return;
intleft = 0;
intright = nums.length - 1;
inti = left;
while (i <= right) {
if (nums[i] == 0) {
swap(nums, i++, left++);
 } elseif (nums[i] == 2) {
swap(nums, i, right--);
 } else {
i++;
 }
 }
 }

privatevoidswap(int[] nums, inti, intj) {
inttmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
 }


publicstaticvoidmain(String[] args) {
SortColors75sc = newSortColors75();

int[] nums = newint[]{
5, 8, 1, 3, 7, 2, 9, 4, 1, 1, 0
 };

System.out.println(Arrays.toString(nums));
sc.sortColors(nums);
System.out.println(Arrays.toString(nums));

nums = newint[]{
1, 1, 0, 2, 0, 1, 2, 0, 1, 1, 1, 2
 };

System.out.println(Arrays.toString(nums));
sc.sortColors(nums);
System.out.println(Arrays.toString(nums));
 }
}