一、860 柠檬水找零

题目解析：注意一开始是没有零钱，也只会取5 10 20三类数字，因此从这3类数字出发，去判断。

1 如果是5元，那么直接收，five++；

2 如果是10元，那么需要five--，ten++，但是如果five本身就小于0，那么不可能找零成功，返回false；

3 如果是20元，这里需要看有多少种零钱找零方式：1）10元+5元组合； 2）5+5+5组合；首先需要消耗10元，因为5元可以兑换零钱方式更多，先把兑换方式少的减掉。

bool lemonadeChange(int* bills, int billsSize) {int five = 0, ten = 0, twenty = 0;for(int i = 0; i < billsSize; i++){if(bills[i] == 5){five++;}else if(bills[i] == 10){ten++;if(five > 0){five--;}else{//five没有 无法找零return false;}}else{if(ten > 0 && five > 0){ten--;five--;}else if(five >= 3){five -=3;}else{return false;}}}return true;
}

二、406 根据身高重建队列

题目要求重建队列，保证身高和前面的人数两个方面，需要从两个方面去拆分排序。

1 先考虑身高，从高到低，如果身高相同，那么k（人数）少的排前面。

2 排序k（人数）因为身高已经排序成功，所以需要考虑的是k值，把k插入对应的位置，这样people身高是排序的，只需要考虑k的位置，就满足了两个条件。

/*** Return an array of arrays of size *returnSize.* The sizes of the arrays are returned as *returnColumnSizes array.* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().*/
void insert(int**arr, int size, int pos, int* val)
{int** tmp = (int**)malloc(sizeof(int*) * size);int index = 0;for(int i = pos; i < size; i++){tmp[index] = malloc(sizeof(int) * 2);tmp[index][0] = arr[i][0];tmp[index][1] = arr[i][1];index++;}//改值arr[pos][0] = val[0];arr[pos][1] = val[1];index = 0;for(int i = pos + 1; i < size; i++){arr[i][0] = tmp[index][0];arr[i][1] = tmp[index][1];index++;}// for(int i = 0; i < size; i++)// {//     printf("val = %d %d\n", arr[i][0], arr[i][1]);// }
}
void swap(int* a, int* b)
{int tmp_1 = a[0];int tmp_2 = a[1];a[0] = b[0];a[1] = b[1];b[0] = tmp_1;b[1] = tmp_2;
}
int** reconstructQueue(int** people, int peopleSize, int* peopleColSize, int* returnSize, int** returnColumnSizes) {*returnSize = peopleSize;*returnColumnSizes = (int*)malloc(peopleSize * sizeof(int));//先排列身高for(int i = 0; i < peopleSize - 1; i++){for(int j = i + 1; j < peopleSize; j++){if(people[j][0] == people[i][0]){if(people[j][1] < people[i][1]){// int tmp_1 = people[j][0];// int tmp_2 = people[j][1];// people[j][0] = people[i][0];// people[j][1] = people[i][1];// people[i][0] = tmp_1;// people[i][1] = tmp_2;swap(people[j], people[i]);}}else if(people[j][0] > people[i][0]){// int tmp_1 = people[j][0];// int tmp_2 = people[j][1];// people[j][0] = people[i][0];// people[j][1] = people[i][1];// people[i][0] = tmp_1;// people[i][1] = tmp_2;swap(people[j], people[i]);}}}//再查看 k值int** queue = (int**)malloc(sizeof(int*) * peopleSize);for(int i = 0; i < peopleSize; i++){queue[i] = (int*)malloc(sizeof(int) * 2);queue[i][0] = people[i][0];queue[i][1] = people[i][1];}for(int i = 0; i < peopleSize; i++){(*returnColumnSizes)[i] = 2;//printf("%d -> pepeple %d %d\n", i, people[i][0], people[i][1]);insert(queue, peopleSize, people[i][1], people[i]);}return queue;
}

三、452. 用最少数量的箭引爆气球

题目分析：用最少的箭引爆气球，主要是考虑重叠情况，如果重叠区间越多，那么使用的箭数量越少，利用排序 + 贪心算法：

void swap(int* a, int* b)
{int tmp_1 = a[0];int tmp_2 = a[1];a[0] = b[0];a[1] = b[1];b[0] = tmp_1;b[1] = tmp_2;
}int findMinArrowShots(int** points, int pointsSize, int* pointsColSize){*pointsColSize = 2;if(pointsSize == 0){return 0;}int result = 1;//按照左边界排序for(int i = 0; i < pointsSize - 1; i++){for(int j = i + 1; j < pointsSize; j++){if(points[j][0] < points[i][0]){swap(points[i], points[j]);}} }for(int i = 1; i < pointsSize; i++){if(points[i][0] > points[i -1][1]){result++;}else{points[i][1] = points[i][1] < points[i -1][1]? points[i][1] : points[i - 1][1];//最小右边界}}return result;
}

上面的code在LeetCode上面会超时，分析了下，应该是排序算法导致的，因此参考快排的算法：

void swap(int* a, int* b)
{int tmp_1 = a[0];int tmp_2 = a[1];a[0] = b[0];a[1] = b[1];b[0] = tmp_1;b[1] = tmp_2;
}
int Partition(int **points, int low, int high){     // 从小到大排序int base = points[low][0];int index = points[low][1];int left,right;while(low < high){while(low < high && points[high][0] >= base){high--;}left = points[low][0];right = points[low][1];points[low][0] = points[high][0];points[low][1] = points[high][1];points[high][0] = left;points[high][1] = right;        while(low < high && points[low][0] <= base){low++;}left = points[low][0];right = points[low][1];points[low][0] = points[high][0];points[low][1] = points[high][1];points[high][0] = left;points[high][1] = right;}return low;
}
void QuickSort(int **points, int low, int high){if(low < high){int base = Partition(points,low,high);QuickSort(points,low,base - 1);QuickSort(points, base + 1, high);}
}int findMinArrowShots(int** points, int pointsSize, int* pointsColSize){*pointsColSize = 2;if(pointsSize == 0){return 0;}int result = 1;//按照左边界排序// for(int i = 0; i < pointsSize - 1; i++)// {//     for(int j = i + 1; j < pointsSize; j++)//     {//         if(points[j][0] < points[i][0])//         {//            swap(points[i], points[j]);//         }//     } // }QuickSort(points, 0 , pointsSize - 1);//左闭右闭区间for(int i = 1; i < pointsSize; i++){if(points[i][0] > points[i -1][1])//如果当前的左边超过了上一个的右边边界，不重叠{result++;}else//找到重叠区间最小右边界{points[i][1] = points[i][1] < points[i -1][1]? points[i][1] : points[i - 1][1];//最小右边界}}return result;
}