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?
思路： 1. 类似快排，走两遍（v=1, 分出0；v = 2，分出1）。
void partition(int A[], int n, int v) {
int start = 0, end = n-1;
while(start < end) {
while(start < end && A[start] < v) ++start;
while(start < end && A[end] >= v) --end;
int tem = A[start];
A[start++] = A[end];
A[end--] = tem;
}
}
class Solution {
public:
void sortColors(int A[], int n) {
partition(A, n, 1);
partition(A, n, 2);
}
};
2. 计数排序。计数与重写。
class Solution {
public:
void sortColors(int A[], int n) {
int count[3] = {0};
for(int i = 0; i < n; ++i) count[A[i]]++;
int id = 0;
for(int i = 0; i < 3; ++i)
for(int j = 0; j < count[i]; ++j)
A[id++] = i;
}
};
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
For example, If n = 4 and k = 2, a solution is:
[ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4], ]
思路：递归，每层从前往后逐步取元素。
void combination(int k, int num, int begin, int end, vector<int> & vec2, vector<vector<int> > &vec) {
if(num == k) {
vec.push_back(vec2);
return;
}
for(int i = begin; i <= end; ++i) {
vec2.push_back(i);
combination(k, num+1, i+1, end, vec2, vec);
vec2.pop_back();
}
}
class Solution {
public:
vector<vector<int> > combine(int n, int k) {
vector<int> vec2;
vector<vector<int> > vec;
combination(k, 0, 1, n, vec2, vec);
return vec;
}
};
52. Sort Colors && Combinations
原文：http://www.cnblogs.com/liyangguang1988/p/3954256.html