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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.
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
https://oj.leetcode.com/problems/maximum-subarray/
思路1:Kadane算法,复杂度O(n)。
思路2:分治。对于每次递归,将数组分半,最大和可能存在
注意点:注意负数的处理,此题最大值为负数时依然返回最大的负数,也有题目负数时要求返回0。注意两种情况下最大值的初始化等细节的区别。
思路1代码:
public class Solution { public int maxSubArray(int[] A) { int n = A.length; int i; int maxSum = A[0]; int thisSum = 0; for (i = 0; i < n; i++) { thisSum += A[i]; if (thisSum > maxSum) maxSum = thisSum; if (thisSum < 0) thisSum = 0; } return maxSum; } public static void main(String[] args) { System.out.println(new Solution().maxSubArray(new int[] { -2, 1, -3, 4, -1, 2, 1, -5, 4 })); } }
思路2代码:
public class Solution { public int maxSubArray(int[] A) { return maxSub(A, 0, A.length - 1); } private int maxSub(int[] a, int left, int right) { if (left == right) return a[left]; int mid = (left + right) / 2; int maxLeft = maxSub(a, left, mid); int maxRight = maxSub(a, mid + 1, right); int leftHalf = 0, leftHalfMax = Integer.MIN_VALUE; int rightHalf = 0, rightHalfMax = Integer.MIN_VALUE; for (int i = mid; i >= left; i--) { leftHalf += a[i]; if (leftHalf > leftHalfMax) leftHalfMax = leftHalf; } for (int i = mid + 1; i <= right; i++) { rightHalf += a[i]; if (rightHalf > rightHalfMax) rightHalfMax = rightHalf; } return Math.max(Math.max(maxLeft, maxRight), (leftHalfMax + rightHalfMax)); } public static void main(String[] args) { System.out.println(new Solution().maxSubArray(new int[] { -2, -1, -3, -2, -5 })); } }参考:
Data Structures and Algorithm Analysis in C
[leetcode] Maximum Subarray,布布扣,bubuko.com
标签:style class blog code java http
原文地址:http://www.cnblogs.com/jdflyfly/p/3810766.html