53. Maximum Subarray

Given 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.

Solution1:

class Solution {
    public int maxSubArray(int[] nums) {
        int currsum = 0;
        int res = Integer.MIN_VALUE;//注意不能初始化为0，因为数组元素可能是负数，所以用Integer.MIN_VALUE，如果有需要，也有Integer.MAX_VALUE.
        for(int num: nums){
            currsum = Math.max(currsum+num,num);
            res = Math.max(res,currsum);
        }
        return res;
    }
}

如果当前sum比当前数组元素小，则给当前sum赋值当前数组元素，太巧妙了（不是，是我笨）。

然后上次循环的result和当前的sum比较取大的即可。

 

二刷，dp真是一种大道至简的做法。

 

 

class Solution {
    public int maxSubArray(int[] nums) {
        int res = nums[0];
        int curmax = 0;
        for(int i = 0; i < nums.length; i++){
            curmax = Math.max(nums[i], curmax + nums[i]);
            res = Math.max(curmax, res);
        }
        return res;
    }
}