53. 最大子序和Java实现，几种方法

1.穷举法

class Solution {
    public int maxSubArray(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int max = Integer.MIN_VALUE;
        for (int begin = 0; begin < nums.length; begin++){
            for (int end = begin; end < nums.length; end++){
                int sum = 0;
                for (int i = begin; i <= end; i++){
                    sum += nums[i];
                }
                max = Math.max(max, sum);
            }
        }
        return max;
    }
}

都2021年了，Java使用这个方法最后几个测试用例提交会超出时间限制。

2.对暴力穷举法的优化：从begin加到end的sum，下一次计算sum的时候可以直接用上次计算过的sum加上nums[end]即可.

class Solution {
    public int maxSubArray(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int max = Integer.MIN_VALUE;
        for (int begin = 0; begin < nums.length; begin++){
            int sum = 0;
            for (int end = begin; end < nums.length; end++){
                sum += nums[end];
                max = Math.max(max, sum);
            }
        }
        return max;
    }
}

3.分治法

class Solution {
    public int maxSubArray(int[] nums){
        return maxSubArray(nums, 0, nums.length);
    }
    public int maxSubArray(int[] nums, int begin, int end) {
        if (end - begin < 2) return nums[begin];
        int mid = (begin + end) >> 1;
        int leftMax = Integer.MIN_VALUE;
        int leftSum = 0;
        
        //中间两个for循环是结果子序列被mid分隔
        for (int i = mid - 1; i >= begin; i--){
            leftSum += nums[i];
            leftMax = Math.max(leftMax, leftSum);
        }
        
        int rightMax = Integer.MIN_VALUE;
        int rightSum = 0;
        for (int i = mid; i < end; i++){
            rightSum += nums[i];
            rightMax = Math.max(rightMax, rightSum);
        }
        
        return Math.max(leftMax + rightMax, Math.max(
                maxSubArray(nums, begin, mid),//最大的子序列在左边的情况
                maxSubArray(nums, mid, end) //最大的子序列在右边的情况
        ));
    }
}