第6章：LeetCode--数组（冒泡排序、快速排序）

11. Container With Most Water

class Solution {
public:
    int maxArea(vector<int>& height) {
        int maxarea = 0;
        int l = 0, r = height.size()-1;
        if(height.size()<2) return maxarea;

        while(l<r){
            maxarea = max(maxarea, min(height[l],height[r])*(r-l));
            if(height[l]<height[r])
                l++;
            else
                r--;
        }

        return maxarea;
    }
};
//Brute Force --Time Limit Exceed
/*class Solution {
public:
    int maxArea(vector<int>& height) {
        int maxarea = 0;
        if(height.size()<2) return maxarea;
        for(int i=0; i<height.size()-1; i++){
            for(int j=i+1; j<height.size(); j++)
                maxarea = max(maxarea, min(height[i],height[j])*(j-i));
        }
        return maxarea;
    }
};*/

16. 3Sum Closest //最接近target的 三数和

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        if(nums.size()<3) return 0;
        if(nums.size() == 3) return nums[0]+nums[1]+nums[2];
        sort(nums.begin(), nums.end());
        //int diff = abs(nums[0]+nums[1]+nums[2]-target);
        int diff = INT_MAX;
        //int sum = 0, front, end, ret=diff;
        int sum = 0, front, end, ret=0;
        for(int i=0; i<nums.size(); i++){
            front = i+1;
            end = nums.size()-1;
            while(front<end){
                sum = nums[i]+nums[front]+nums[end];
                if(sum == target) return target;
                if(abs(sum-target)<diff){
                    diff = abs(sum-target);
                    ret = sum;
                }
                (sum > target) ? end-- : front++;
            }
        }
        return ret;
    }
};

49. Group Anagrams 　

Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
Output:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map<string, vector<string>> mp;
        for (string s : strs) { //auto s: strs also ok
            string t = s;
            sort(t.begin(), t.end());
            mp[t].push_back(s);
        }
        vector<vector<string>> anagrams;
        for (auto p : mp) {
            anagrams.push_back(p.second);
        }
        return anagrams;
    }
};

819. Most Common Word

Input:
paragraph = "Bob hit a ball, the hit BALL flew far after it was hit."
banned = ["hit"]
Output: "ball"

class Solution {
public:
    string mostCommonWord(string paragraph, vector<string>& banned) {

        // Map out each banned word
        // Go through each word, if it isn't in the banned list
        // put it in another map and keep track of how many times it appears
		// Keep two variables to maintain the mostCommonWord and how many times it appeared
			// This prevents us from looping thorugh the map at the end to find the most common word

        map<string, int> bannedWords;
        map<string, int> tracker;

        string mostCommonWord = "";
        int mostCommonWordCount = 0;

        for(int x = 0; x < banned.size(); x++)
        {
            bannedWords[banned[x]] = 0;
        }

        for(int x = 0; x < paragraph.length(); x++)
        {
            string temp = "";
            while(x < paragraph.length() && paragraph[x] != ' ')
            {
                if(!isalpha(paragraph[x]))
                {
                    break;
                }

                temp += tolower(paragraph[x]);
                x++;
            }

            if(temp == " " || temp == "")
                continue;

            if(bannedWords.count(temp) != 0)
                continue;
            else
            {
                tracker[temp]++;

                if(mostCommonWordCount < tracker[temp])
                {
                    mostCommonWordCount = tracker[temp];
                    mostCommonWord = temp;
                }
            }

        }

        return mostCommonWord;
    }
};

Question1？？？:

Input:
inputStr = awaglknagawunagwkwagl
num = 4

Output:
{wagl, aglk, glkn, lkna, knag, gawu, awun, wuna, unag, nagw, agwk, kwag}
Explanation:
Substrings in order are: wagl, aglk, glkn, lkna, knag, gawu, awun, wuna, unag, nagw, agwk, kwag, wagl
"wagl" is repeated twice, but is included in the output once.

class Solution {
public:
    vector<string> numKLenSubstrNoRepeats(string S, int K) {
        vector<string>ans;
        int cnt = 0, ,i=0, j;
        while(i < S.size() && j<k) {
            if(S[i] == S[j]){
　　　　　　　　　　j++;
　　　　　　　}else{
　　　　　　　　　　i = i-j+1;
　　　　　　　　　　j = 0;
　　　　　　　}
            if (j == K){
　　　　　　 　　ans.push_back(S.substr(i-K, K));
　　　　　　　　 i = i-j+1;
　　　　　　　　 j = 0;
　　　　　　　}
        }
        return ans;
    }
};　　

数组排序

https://www.cnblogs.com/taotingkai/p/6214367.html

https://baike.baidu.com/item/%E5%BF%AB%E9%80%9F%E6%8E%92%E5%BA%8F%E7%AE%97%E6%B3%95/369842?fromtitle=%E5%BF%AB%E9%80%9F%E6%8E%92%E5%BA%8F&fromid=2084344&fr=aladdin#3_5

/* 冒泡排序 */
/* 1. 从当前元素起，向后依次比较每一对相邻元素，若逆序则交换 */ （每次把最大值放后面）
/* 2. 对所有元素均重复以上步骤，直至最后一个元素 */
void bubbleSort (int arr[], int len) {
    int temp;
    int i, j;
    for (i=0; i<len-1; i++) /* 外循环为排序趟数，len个数进行len-1趟 */
        for (j=0; j<len-1-i; j++) { /* 内循环为每趟比较的次数，第i趟比较len-i次 */
            if (arr[j] > arr[j+1]) { /* 相邻元素比较，若逆序则交换（升序为左大于右，降序反之） */
                temp = arr[j];
                arr[j] = arr[j+1];
                arr[j+1] = temp;
            }
        }
}
/* 每次选出最小的 */
void bubbleSort (int arr[], int len) {
    int temp;
    int i, j;
    for (i=0; i<len-1; i++) /* 外循环为排序趟数，len个数进行len-1趟 */
        for (j=i+1; j<len-1; j++) { /* 内循环为每趟比较的次数，第i趟比较len-i次 */
            if (arr[i] > arr[j]) { /* 相邻元素比较，若逆序则交换（升序为左大于右，降序反之） */
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
}

/*快速排序*/
#include <iostream>
using namespace std;
void Qsort(int arr[], int low, int high){
    if (high <= low) return;
    int i = low;
    int j = high + 1;
    int key = arr[low];
    while (true)
    {
        /*从左向右找比key大的值*/
        while (arr[++i] < key)
        {
            if (i == high){
                break;
            }
        }
        /*从右向左找比key小的值*/
        while (arr[--j] > key)
        {
            if (j == low){
                break;
            }
        }
        if (i >= j) break;
        /*交换i,j对应的值*/
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }
    /*中枢值与j对应值交换*/
    int temp = arr[low];
    arr[low] = arr[j];
    arr[j] = temp;
    Qsort(arr, low, j - 1);
    Qsort(arr, j + 1, high);
}

int main()
{
    int a[] = {6, 2, 7, 3, 8, 9};
    Qsort(a, 0, sizeof(a) / sizeof(a[0]) - 1);/*这里原文第三个参数要减1否则内存越界*/
    for(int i = 0; i < sizeof(a) / sizeof(a[0]); i++)
    {
        cout << a[i] << "";
    }
    return 0;
}/*参考数据结构p274(清华大学出版社，严蔚敏)*/

求一个数组里面连续子集的和的最大值，比如[1,2,-4,7,8,-2,4]=>[7,8,-2,4]

思路：每次求出max(sub(i-1), arr[i])，看那个大取那个。

int arrMaxsub(int [] arr, int len){
　　int maxsum = arr[0];

    for(int i=1; i<len; i++){
        if(maxsum+arr[i] <arr[i])
            maxsum = arr[i];
        else
            maxsum = maxsum+arr[i];
    }      
    return maxsum;
}