Intuition

The basic intuition comes from sorting.

Approach

In the naive approach, we can sort the array using inbuilt sorting function. The time complexity will be O(N*log(N)).

• Optimize: Since we are sorting only three numbers, we can use the concept of counting sort. Keep track of number of zeros and number of ones in the array.

Complexity

• Time complexity: O(N)

• Space complexity: O(1)

Code

class Solution {
    public void sortColors(int[] nums) {
        int countZero = 0;
        int countOne  =  0;
        for(int num: nums){
            switch(num){
                case 0:
                    countZero++;
                    break;
                case 1:
                    countOne++;
            }
        }
        int currentIndex = -1;
        while(0<countZero--){
            nums[++currentIndex] = 0;
            // countZero--;
        }
        while(0<countOne--){
            nums[++currentIndex] = 1;
            // countOne--;
        }
        while(currentIndex<nums.length-1){
            nums[++currentIndex] = 2;
        }
    }
}
class Solution {
    public void sortColors(int[] nums) {
        int countZero = 0;
        int countOne  =  0;
        for(int num: nums){
            switch(num){
                case 0:
                    countZero++;
                    break;
                case 1:
                    countOne++;
            }
        }
        int currentIndex = -1;
        while(0<countZero--){
            nums[++currentIndex] = 0;
            // countZero--;
        }
        while(0<countOne--){
            nums[++currentIndex] = 1;
            // countOne--;
        }
        while(currentIndex<nums.length-1){
            nums[++currentIndex] = 2;
        }
    }
}
class Solution {
    public void sortColors(int[] nums) {
        int countZero = 0;
        int countOne  =  0;
        for(int num: nums){
            switch(num){
                case 0:
                    countZero++;
                    break;
                case 1:
                    countOne++;
            }
        }
        int currentIndex = -1;
        while(0<countZero--){
            nums[++currentIndex] = 0;
            // countZero--;
        }
        while(0<countOne--){
            nums[++currentIndex] = 1;
            // countOne--;
        }
        while(currentIndex<nums.length-1){
            nums[++currentIndex] = 2;
        }
    }
}

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GitHub repo for more solutions: Git
Leetcode profile: Leetcode: devn007

原文链接：Leetcode 75. Sort Colors