标签：div blog nbsp hat mcs i++ element ref code
https://leetcode.com/problems/maximum-subarray/description/
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.
public class Solution { public int maxSubArray(int[] nums) { // DP // Time O（N） Space O(1) int maxLocal = nums[0]; int global = nums[0]; for (int i = 1; i < nums.length; ++i){ maxLocal = Math.max(nums[i], nums[i] + maxLocal); global = Math.max(global, maxLocal); } return global; } }
Sol 2:
(Unsolved -- don‘t know what cur_min is doing here...)
public class Solution { public int maxSubArray(int[] nums) { // Time O（n） Space O(n) return mcss(nums, 0, nums.length); } // find the max sum of the continuous array public static int mcss(int[] nums, int begin, int end){ final int n = end - begin; int[] sum = new int[n + 1]; // the sum of n elements in the front int result = Integer.MIN_VALUE; int cur_min = sum[0]; for (int i = 1; i <= n; i++){ sum[i] = sum[i - 1] + nums[begin + i - 1]; } for (int i = 1; i <= n; i++){ result = Math.max(result, sum[i] - cur_min); cur_min = Math.min(cur_min, sum[i]); } return result; } }
标签：div blog nbsp hat mcs i++ element ref code
原文地址：http://www.cnblogs.com/prmlab/p/7295514.html
