53. Maximum SubArray(divide and conquer)
2017-09-10 21:27
399 查看
题目描述
Find the contiguous subarray within an array (containing at least one number) which has the largest sum.For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.
算法分析
题目的要求是从数组中找出和最大的子数组,算法的思想是先找到一个非负数(如果没有,也就说都是负数的话,第一个数就是和最大的子数组)。当找到第一个非负数开始,记为当前的最大和ans,继续扫,如果sum变小了,那么ans不变,如果sum变得小于0了,那么ans包括的即是当前和最大的子数组A,跳过这些使sum变为负的子数组,寻找另一个以非负数开始的子数组B,当然,如果sum(子数组B)大于sum(子数组A),则ans为子数组B的和。算法设计
class Solution { public: int maxSubArray(vector<int> nums) { int len = nums.size(); int ans = nums[0], sum = 0, index = 0; while(index < len) { sum += nums[index]; ans = max(sum, ans); if(sum < 0 ) { sum = 0; } index++; } return ans; } int max(int a, int b) { return (a>b)? a:b; } };
相关文章推荐
- 分治法(divide-and-conquer)基础(
- [算法]分治算法(Divide and Conquer)
- 分治 Divide and Conquer
- Divide and Conquer:River Hopscotch(POJ 3258)
- 215. Kth Largest Element in an Array(divide and conquer)
- Leetcode Algorithms - Divide and Conquer : 53. Maximum Subarray
- Divide and conquer:Median(POJ 3579)
- 分治法——主定理(Divide and Conquer - The Master Theorem)
- 算法上机题目mergesort,priority queue,Quicksort,divide and conquer
- 分治 Divide and Conquer 局部最小值 local optimal 棋盘问题
- [Algorithm]分治法 Divide and Conquer 与 主定理 Master Theorem
- arraymethodDivide and Conquer
- algorithm:divide and conquer
- 327. Count of Range Sum(Divide and Conquer)
- Factorial Trailing Zeroes (Divide-and-Conquer)
- 分治 Divide and Conquer
- Leetcode Algorithms - Divide and Conquer:169. Majority Element
- divide and conquer
- 【算法】【Divide and conquer】Median of Two Sorted Arrays
- [TreeDivideAndConquer]点分治