LeetCode-Maximum Subarray
来源:互联网 发布:智能网络电视排行榜 编辑:程序博客网 时间:2024/05/17 01:40
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
.
public int maxSubArray(int[] nums) { if (nums == null || nums.length == 0) return 0; int maxSoFar = nums[0]; int maxEndingHere = nums[0]; for (int i = 1; i < nums.length; i++) { maxEndingHere = Math.max(nums[i], maxEndingHere+nums[i]); maxSoFar = Math.max(maxSoFar, maxEndingHere); } return maxSoFar; }
解法非常直观,maxEndingHere是指以sum(0, i),即起点为o,终点为i这一段的最大值。而此时对于第i个元素有两个选择,要么加入之前的和之中,即maxEndingHere+nums[i],否则不用之前的和。取最大值即可。公式表示为:
sum[i] = max(nums[i], sum[i-1] + nums[i+1])
result = max(result, sum[i])
显然nums[i]与nums[i]+sum[i-1]孰大孰小,完全取决于sum[i-1]的正负,所以代码也可以这样写:
public int maxSubArray(int[] A) { if (A == null || A.length == 0) { return 0; } int max = A[0]; int curSum = A[0]; for (int i = 1; i < A.length; i++) { if (curSum < 0) { curSum = A[i]; } else { curSum += A[i]; } if (max < curSum) { max = curSum; } } return max; }
这个算法还是比较经典的,好多算法也用到了这个特性,比如,maxEndingHere或者maxBeginingHere。
0 0
- 【LeetCode】Maximum Subarray 和 Maximum Product Subarray
- LeetCode: Maximum Subarray
- LeetCode Maximum Subarray
- [Leetcode] Maximum Subarray
- LeetCode: Maximum Subarray
- leetcode 25: Maximum Subarray
- [LeetCode] Maximum Subarray
- [Leetcode] Maximum Subarray
- [LeetCode]Maximum Subarray
- [leetcode]Maximum Subarray
- LeetCode-Maximum Subarray
- [leetcode] Maximum Subarray
- LeetCode 45: Maximum Subarray
- LeetCode - Maximum Subarray
- LeetCode:Maximum Subarray
- LeetCode 53: Maximum Subarray
- 【leetcode】Maximum Subarray
- Leetcode Maximum Subarray
- 如何在window下操作和连接sqlite 的db文件
- 算法竞赛入门经典第三章
- Geohash的原理、算法和具体应用探究(优秀)
- Error:无法打开 源文件 “stdafx.h”
- 如何从pdf文件中取出图片
- LeetCode-Maximum Subarray
- 使用AsyncTask运行异步任务
- mysql版本分类
- baseDao的用法
- POJ - 2524 Ubiquitous Religions(简单并查集)
- 【算法】【动态规划】Bus Fare
- java中instanceof用法
- 衡量算法的效率
- 编程软件字体推荐(一)