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;
    }
}