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  • 53. Maximum Subarray (JAVA)

    iven an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

    Example:

    Input: [-2,1,-3,4,-1,2,1,-5,4],
    Output: 6
    Explanation: [4,-1,2,1] has the largest sum = 6.
    

    Follow up:

    If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

    法I: 动态规划法

    class Solution {
        public int maxSubArray(int[] nums) {
            int maxSum = Integer.MIN_VALUE; //注意有负数的时候,不能初始化为0
            int currentSum = Integer.MIN_VALUE;
            for(int i = 0; i < nums.length; i++){
                if(currentSum < 0) currentSum = nums[i];
                else currentSum += nums[i];
                
                if(currentSum > maxSum) maxSum = currentSum; 
            }
            return maxSum;
        }
    }

     法II:分治法

    class Solution {
        public int maxSubArray(int[] nums) {
            return partialMax(nums,0,nums.length-1);
        }
        
        public int partialMax(int[] nums, int start, int end){
            if(start == end) return nums[start];
           
            int mid = start + ((end-start) >> 1);
            int leftMax = partialMax(nums,start, mid);
            int rightMax = partialMax(nums,mid+1,end);
            int maxSum = Math.max(leftMax,rightMax);
            
            int lMidMax = Integer.MIN_VALUE;
            int rMidMax = Integer.MIN_VALUE;
            int current = 0;
            for(int i = mid; i >= start; i--){
                current += nums[i];
                if(current > lMidMax) lMidMax = current;
            }
            current = 0;
            for(int i = mid+1; i <= end; i++){
                current += nums[i];
                if(current > rMidMax) rMidMax = current;
            }
            if(lMidMax > 0 && rMidMax > 0) maxSum = Math.max(lMidMax + rMidMax,maxSum);
            else if(lMidMax > rMidMax) maxSum = Math.max(lMidMax,maxSum);
            else maxSum = Math.max(rMidMax,maxSum);
            return maxSum;
        }
    }
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  • 原文地址:https://www.cnblogs.com/qionglouyuyu/p/10901168.html
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