52. Sort Colors && Combinations

最新推荐文章于 2024-02-04 22:04:59 发布

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最新推荐文章于 2024-02-04 22:04:59 发布

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原文链接：http://www.cnblogs.com/liyangguang1988/p/3954256.html

本文深入探讨了排序颜色和组合问题的解决方法，通过详细分析两种算法的思路和实现，帮助开发者理解如何使用计数排序和递归来解决实际问题。

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Sort Colors

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.

click to show follow up.

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;
    }
};

Combinations

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;
    }
};

转载于:https://www.cnblogs.com/liyangguang1988/p/3954256.html