标签：ace turn for cti 思路 统计 void i++ pos
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.
思路：
首先扫描一遍，统计不同元素个数。
类似于two pointers概念。
void sortColors(vector<int>& nums) { if (nums.size() <= 1)return; vector<int>temp = nums; int zero = 0, one = 0, two = 0, index0 = 0, index1 = 0, index2 = 0; for (int i = 0; i < nums.size();i++) { if (nums[i] == 0)zero++; if (nums[i] == 1)one++; if (nums[i] == 2)two++; } index1 = zero, index2 = zero + one; for (int i = 0; i < temp.size(); i++) { if (temp[i] == 0)nums[index0++] = 0; if (temp[i] == 1)nums[index1++] = 1; if (temp[i] == 2)nums[index2++] = 2; } }
标签：ace turn for cti 思路 统计 void i++ pos
原文地址：http://www.cnblogs.com/hellowooorld/p/6932066.html
