greedy: 55. Jump Game \ 455. Assign Cookies

2/10/2017来源:ASP.NET技巧人气:1602

Jump Game 题目描述 代码实现 Assign Cookies 题目描述 代码实现

55. Jump Game

题目描述

Given an array of non-negative integers, you are initially positioned at the first index of the array. Each element in the array rePResents your maximum jump length at that position. Determine if you are able to reach the last index. For example: A = [2,3,1,1,4], return true. A = [3,2,1,0,4], return false.

代码实现

法一:

// 复杂度O(n^2)的做法,导致超时了。 class Solution { public: bool canJump(vector<int>& nums) { int nums_len = nums.size() - 1; vector<bool> flag(nums_len, false); int stt = 0; int jmp = 0; flag[0] = true; cout << nums_len << endl; for(stt = 0; stt <= nums_len; stt++) { if(flag[stt]) { int tmp = nums[stt] + stt; for(int in_range = tmp; in_range >= stt; in_range--) { if(in_range >= nums_len) return true; flag[in_range] = true; // cout << in_range << " " << flag[in_range] << " " << stt << endl; } } } return false; } };

法二: 修改第二个进入循环的条件,没有必要重复设置可以跳转的标志位。这样做了以后可以击败95%的c++代码。

class Solution { public: bool canJump(vector<int>& nums) { int nums_len = nums.size() - 1; vector<bool> flag(nums_len, false); int stt = 0; int jmp = 0; int out_range = -1; flag[0] = true; cout << nums_len << endl; for(stt = 0; stt <= nums_len; stt++) { if(flag[stt]) { int tmp = nums[stt] + stt; if(tmp > out_range) { for(int in_range = tmp; in_range > out_range; in_range--) { if(in_range >= nums_len) return true; flag[in_range] = true; // cout << in_range << " " << flag[in_range] << " " << stt << endl; } out_range = tmp; } } } return false; } };

把上面的代码做些简化设计,可以得到:

class Solution { public: bool canJump(vector<int>& nums) { int nums_len = nums.size(); int i = 0, maxreach = 0; for (; i < nums_len && i <= maxreach && maxreach < nums_len - 1; ++i) maxreach = max(maxreach,i+nums[i]); return maxreach>=nums_len-1; } };

这种做法比较简约,但是效率比较低。因为它需要对所有的情况调用max函数,遇到较大的数组的时候就会比较耗时。这种做法不如我在法二中做的剪枝那么有效。

455. Assign Cookies

题目描述

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.

代码实现

这种做法击败了99.24%的做法,思路比较简单。就是先排序在比较。比较的时候,如果比较到s的索引为s_stt,s_stt之前的都没有必要在下一次比较。

class Solution { public: int findContentChildren(vector<int>& g, vector<int>& s) { int content_num = 0; int g_len = g.size(); int s_len = s.size(); sort(g.begin(), g.end()); sort(s.begin(), s.end()); int s_stt = 0; for(int i = 0; i < g_len; i++) { for(int j = s_stt; j < s_len; j++) { if(g[i] <= s[j]) { content_num++; s_stt = j+1; break; } } } return content_num; } };