Three-color sorting

Main topic:

Give you a red, white, blue array of three colors, ordered according to the Order of red, white, and blue, Leetcode link: https://leetcode.com/problems/sort-colors/

Idea 1:

Sort the entire array directly, time complexity O (NLOGN)

Idea 2:

Count sorting method, use a hash-like array to record the number of each color, and then sort, but need to traverse the original array two times

Idea 3:

Using three pointers, P1 represents the dividing line between red and white, p2 represents the dividing line between white and blue, and I represents the current element

That is, 0~p1-1 is red, p1~i-1 is white, p2+1~n-1 means blue.

1) If the current element is red, it is exchanged with the element pointed to by P1, because the color I refer to after the interchange is white, I traverse the next element directly

2) If the current element is blue, it is exchanged with the element pointed to by the P2, because the color I refer to after the interchange may be white, it may be red, so I need to rewind I

The implementation code is as follows: note I should be between [P1,P2]

classsolution{ Public:    voidSortcolors (vector<int>&nums) {    intn =nums.size (); if(N <=1)    {    return; }     intP1 =0; intP2 = n-1; //Notice the cyclic condition of I     for(intI=0; i<=p2; ++i) {if(Nums[i] = =0) {swap (NUMS[P1], nums[i]); ++P1; }    Else if(Nums[i] = =2) {swap (NUMS[P2], nums[i]); --P2; --i; }    }    }Private:    voidSwapint&a,int&b) {intTMP =A; A=b; b=tmp; }};

Three-color sorting