贪心算法3

• 全局思考：总油量减去总消耗大于等于零那么一定可以跑完一圈,
• 局部贪心：累加每个站的净胜油量,如果<0,则在此之前(包括该站)都不是起始位置,从下一个位置开始寻找

class Solution {
public:int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {int cursum = 0, totalsum = 0;int start = 0;for (int i = 0; i < gas.size(); i++) {cursum += gas[i] - cost[i];//累加油量净剩量totalsum += gas[i] - cost[i];if (cursum < 0) {//i位置之前一定不满足start = i + 1;//从i下一个位置开始重新寻找cursum = 0;}}if (totalsum < 0)//总油量大于消耗的，所以肯定不能跑一圈return -1;return start;}
};

两次遍历

1. 从左到右，考虑右边比左边大的情况
2. 从右到左，考虑左边比右边大的情况

class Solution {
public:int candy(vector<int>& ratings) {int ret = 0;vector<int> Candy(ratings.size(), 1);for (int i = 1; i < ratings.size(); i++) {if (ratings[i - 1] < ratings[i])Candy[i] = Candy[i - 1] + 1;}for (int i = ratings.size() - 2; i >= 0; i--) {if (ratings[i] > ratings[i + 1])Candy[i] = max(Candy[i], Candy[i + 1] + 1);}for (int num : Candy)ret += num;return ret;}
};

贪心体现在如果支付20美元时，应该优先消耗10美元找零

class Solution {
public:bool lemonadeChange(vector<int>& bills) {int five = 0, ten = 0;for (int b : bills) {if (b == 5) five++;else if (b == 10) five--, ten++;else {if (ten) ten--, five--;else five -= 3;}if (five < 0) return false;}return true;}
};

首先顾一个维度，按照身高排序，之后优先按身高高的people的k来插入，后序插入节点不会影响前面已经插入的节点

class Solution {
public:static bool cmp(const vector<int>& a, const vector<int>& b) {if (a[0] == b[0]) return a[1] < b[1];return a[0] > b[0];}vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {sort(people.begin(), people.end(), cmp);vector<vector<int>> que;for (int i = 0; i < people.size(); i++) {int pos = people[i][1];que.insert(que.begin() + pos, people[i]);}return que;}
};