Triangle Count】的更多相关文章

Given an array of integers, how many three numbers can be found in the array, so that we can build an triangle whose three edges length is the three numbers that we find? Example Given array S = [3,4,6,7], return 3. They are: [3,4,6] [3,6,7] [4,6,7]…
题目: Given an array of integers, how many three numbers can be found in the array, so that we can build an triangle whose three edges length is the three numbers that we find? Example Given array S = [3,4,6,7], return 3. They are: [3,4,6] [3,6,7] [4,6…
Should be "Medium" or even "Easy".. Just with a little Greedy. class Solution { public: /** * @param S: A list of integers * @return: An integer */ int triangleCount(vector<int> &S) { int n = S.size(); ) ; sort(S.begin(), S.e…
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. For example, given the following triangle [ [2], [3,4], [6,5,7], [4,1,8,3] ] The minimum path sum from top to bottom is 11 (i…
Yet Another Source Code for LintCode Current Status : 232AC / 289ALL in Language C++, Up to date (2016-02-10) For more problems and solutions, you can see my LintCode repository. I'll keep updating for full summary and better solutions. See cnblogs t…
------------------------------------------------------------第七周:Follow up question 1,寻找峰值 寻找峰值 描述 笔记 数据 评测 你给出一个整数数组(size为n),其具有以下特点: 相邻位置的数字是不同的 A[0] < A[1] 并且 A[n - 2] > A[n - 1] 假定P是峰值的位置则满足A[P] > A[P-1]且A[P] > A[P+1],返回数组中任意一个峰值的位置. 注意事项 数…
1. PageRank http://blog.csdn.net/hguisu/article/details/7996185 2. Connected Components 3. Triangle Counting 例子: users.txt 1,BarackObama,Barack Obama 2,ladygaga,Goddess of Love 3,jeresig,John Resig 4,justinbieber,Justin Bieber 6,matei_zaharia,Matei Z…
原题链接在这里:https://leetcode.com/problems/3sum-smaller/ 题目: Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target. For e…
3Sum Smaller 要点:类似的题还有lintcode的triangle count:https://github.com/delbao/onlineJudge/blob/master/lintcode/109.triangle-count.java i,j,k的规律:i:最外层,每走一步,j要重新来过,j第二层,起始为i+1,k初始取决于题目.主要利用了: [j,k]之间的增序:当k满足条件后,顺序之后的也都满足条件,这时候就可以继续移动j了. 移动j的时候,由于j位置的元素变大了,k在…
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