Leetcode  455

思路：

1、贪心度越低的孩子越容易满足

2、while循环。糖果价值按由小到大匹配贪心度按由小到大排序的孩子，如果最低糖果价值不能满足最低贪心度的孩子，则舍弃该糖果，直到匹配到等价的。

这样时间复杂度是O(nlogn)，快速排序占的时间比较高。实现代码如下：

/*
题目：
Assume you are an awesome parent and want to give your children some cookies.
But, you should give each child at most one cookie.
Each child i has a greed factor gi, which is the minimum size of a cookie that the child will be content with;
and each cookie j has a size sj.
If sj >= gi, we can assign the cookie j to the child i, and the child i will be content.
Your goal is to maximize the number of your content children and output the maximum number.

Note:
You may assume the greed factor is always positive.
You cannot assign more than one cookie to one child.

Example 1:
Input: [1,2,3], [1,1]

Output: 1

Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3.
And even though you have 2 cookies, since their size is both 1,
you could only make the child whose greed factor is 1 content.
You need to output 1.

Example 2:
Input: [1,2], [1,2,3]

Output: 2

Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2.
You have 3 cookies and their sizes are big enough to gratify all of the children,
You need to output 2.
*/

#include "stdafx.h"
#include<iostream>
#include<algorithm>
//学习一下怎么写宏
#define swap( i, j){int q=0;q=i;i=j;j=q; }
using namespace std;

//复习快速排序
void quickSort(int s[], int l, int r) {
	if (l < r) {
		int middle = (l + r) / 2;
		int i = l - 1;
		int j = r + 1;
		while (true) {
			//找出中间位置左侧大于等于中间位置值的数
			while (s[++i] < s[middle]);
			//找出中间位置右侧大于等于中间位置值的数
			while (s[--j] > s[middle]);
			//当i在j后面，就不需要交换
			if (i >= j) break;
			//调用宏，交换获取的两个位置的值			
			swap(s[i], s[j]);
		}
		//递归调用
		quickSort(s, l, i - 1);
		quickSort(s, j + 1, r);
	}
}



//返回最佳匹配值
int Search(int Child[], int Candy[], int Child_num, int Candy_num) {
	int answer = 0, temp_Child = 0, temp_Candy = 0;
	while (temp_Child < Child_num && temp_Candy < Candy_num) {
		//如果孩子贪心度小于等于糖果价值，则记录
		if (Child[temp_Child] <= Candy[temp_Candy]) {
			answer++;
			temp_Candy++;
			temp_Child++;
		}
		//此时糖果不满足孩子的贪心度，舍弃该糖果
		else {
			temp_Candy++;
		}
	}
	return answer;

}

int main()
{
	//Child_num用于记录孩子的个数，Candy_num记录糖果的个数
	//temp_Child用于记录每一次输入的孩子的贪心度，temp_Candy用于记录每一次输入的糖果价值
	int Child_num = 0, Candy_num = 0, temp_Child = 0, temp_Candy = 0;

	cout << "请输入孩子的个数" << endl;
	cin >> Child_num;
	cout << "请输入糖果的块数" << endl;
	cin >> Candy_num;

	//创建数组，用于存储输入的数据
	int*Child;
	int*Candy;
	Child = (int*)malloc(sizeof(int)*Child_num);
	Candy = (int*)malloc(sizeof(int)*Candy_num);

	cout << "请输入每个孩子的贪心度" << endl;
	for (int i = 0; i < Child_num; i++) {
		cin >> temp_Child;
		Child[i] = temp_Child;
	}

	cout << "请输入每个糖果的价值" << endl;
	for (int i = 0; i < Candy_num; i++) {
		cin >> temp_Candy;
		Candy[i] = temp_Candy;
	}

	//调用乱序函数，将输入的数据打乱，这样可以提高快排的效率  
	int*   a_begin = Child;
	int*   a_end = Child + Child_num;
	random_shuffle(a_begin, a_end);

	a_begin = Candy;
	a_end = Candy + Candy_num;
	random_shuffle(a_begin, a_end);

	//调用快速排序函数  O(nlogn)
	quickSort(Child, 0, Child_num - 1);
	quickSort(Candy, 0, Candy_num - 1);

	int answer = Search(Child, Candy, Child_num, Candy_num);
	cout << "满意值为：" << answer << endl;

	return 0;
}

LeetCode   376  

理解：给定一组序列，如果前者与后者的差值依次在正负整数之间交叉出现（开头值正负皆可），则该序列是交叉序列。输出序列中所含最长的交叉序列中元素的个数。（可以删除序列中的某个位置的值（0也可以）来延长交叉序列的长度）。

时间复杂度控制在O(n)

思路：用贪心算法，如果出现两个连续的同符差值，差值为负则证明此段输入序列是三个递增的数，去掉中间的那个数，差值为正则证明此段序列是三个递减数，则去掉中间的数。如此，如果按例题1中的输入序列，最终保留的输出序列应为[1,17,5,15,5,16,8]，与例子1给出的输出序列不一致，但是结果是对的。而且可以保证是O(n)。

以下为思路3的实现代码：

#include "stdafx.h"
#include<iostream>
#include<vector>
using namespace std;

int Output(vector<int>&receive) {	
	/*
	v.empty()               如果v为空，则返回true，否则返回false
	v.size()                返回v中元素的个数
	v.push_back()           在v的末尾增加一个值为t的元素
	v[n]                    返回v中位置为n的元素
	v1=v2                   把v1中的元素替换为v2中元素的副本
	v1==v2                  如果v1与v2相等，则返回true
	!=,<,<=,>,>=            保持这些操作符惯有含义
	*/
	//标记
	int count = 0;
	vector<int>temp;
	for (int i = 0; i < receive.size()-1; i++) {
		//算出序列差值
		temp.push_back ( receive[i] - receive[i + 1]);
	}
	for (int i = 0; i < temp.size()-1; i++) {
		if ((temp[i] > 0 && temp[i + 1] > 0) || (temp[i] < 0 && temp[i + 1] < 0)) {
			//[1,17,5,10,13,15,10,5,16,8]   
			//[-16,  12，-5，-3，-2,  5,  5，-11, 8] 
			//将三个值的中间值制为零
			receive[i + 1] = 0;
		}
	}
	//我现在不确定vector中值为0的位置会不会被size出来，所以先用temp的长度
	for (int i = 0; i < temp.size() + 1; i++) {
		if (receive[i] != 0) {
			count++;
		}
	}
	return count;
}


int main()
{
	cout << "请输入序列中的元素" << endl;
	cout << "结束输入请输入“999”" << endl;
	//试试用容器存数据，不用数组了
	vector<int> receive;

	//存储每次输入的数据
	int temp = 0;
	
	while (cin >> temp) {
		if (temp == 999)break;
		receive.push_back(temp);
	}
	int output = Output(receive);
	cout << "最长差值列中元素的个数是：" << output << endl;
	system("pause");
    return 0;
}

LeetCode 402

题意：由非负整数构成的字符串，从字符串中取出k个数使得新的数最小。

注意：输入的数字串长度小于10002且大于k；输入的数字串不会以0开头。

思路：将字符串中字符依次入栈，如果当前字符串比栈顶元素小，并且还可以继续删除元素，那么就将栈顶元素删掉，这样会得到最小数字序列。

实现代码如下：

#include "stdafx.h"
#include<iostream>
#include<string>
#include<stack>
using namespace std;

string removeKdigits(string num, int k) {
	int n = k, index=0;
	string answer;
	for (size_t i = 0; i < num.size(); i++) {
		while (answer.size() != 0 && n > 0 && num[i] < answer.back()) {
			n--;
			answer.pop_back();
		}
		answer.push_back(num[i]);
	}
	//去除开头的0
	for (size_t i = 0; i < answer.size(); i++) {
		if (answer[i] != '0') {
			index = i;
			break;
		}
		//此时说明answer里面存的全是0
		else if(answer[i]=='0'&&i==answer.size()-1){
			answer = "0";
		}
	}
	//获取答案
	//string a=s.substr(0,5);       
	//获得字符串s中 从第0位开始的长度为5的字符串
	answer = answer.substr(index, num.size() - k - index);
	if (answer.size()==0) answer.push_back('0');
	return answer;
}


int main()
{
	int k = 0;
	char num[10002];
	cin >> num;
	cin >> k;
	string answer=removeKdigits( num,  k);
        return 0;
}

LeetCode  55

题意：给出一组非负整数，你位于数组索引位置为0的地方，数组中的每个值代表你在当前位置所能向后跳跃的最大“格数”（你可以不跳最大格数）。判断你是否能到达索引位置的最后一位。

LeetCode  45

LeetCode  452