Content

Given an integer array nums, find the subarray with the largest sum, and return its sum.

Example 1:

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

Example 2:

Input: nums = [1]
Output: 1
Explanation: The subarray [1] has the largest sum 1.

Example 3:

Input: nums = [5,4,-1,7,8]
Output: 23
Explanation: The subarray [5,4,-1,7,8] has the largest sum 23.

Constraints:

• 1 <= nums.length <= 105
• -104 <= nums[i] <= 104

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.

Solution

1. 动态规划

Java

class Solution {
    public int maxSubArray(int[] nums) {
        // 1 <= nums.length <= 10⁵
        // -10⁴ <= nums[i] <= 10⁴
        int n = nums.length;
        // dp[i]表示以下标i结尾的求和最大的子数组的求和的值
        int[] dp = new int[n];
        dp[0] = nums[0];
        int max = dp[0];
        for (int i = 1; i < n; i++) {
            if (dp[i - 1] > 0) {
                dp[i] = dp[i - 1] + nums[i];
            } else {
                dp[i] = nums[i];
            }
            max = Math.max(max, dp[i]);
        }
        return max;
    }
}

2. 分治法

Java

class Solution {
    public int maxSubArray(int[] nums) {
        // 1 <= nums.length <= 10⁵
        // -10⁴ <= nums[i] <= 10⁴
        Status status = get(nums, 0, nums.length - 1);
        return status.mSum;
    }

    public Status get(int[] nums, int l, int r) {
        if (l > r) {
            return new Status(0, 0, 0, 0);
        } else if (l == r) {
            return new Status(nums[l], nums[l], nums[l], nums[l]);
        } else {
            int mid = l + r >> 1;
            Status lStatus = get(nums, l, mid);
            Status rStatus = get(nums, mid + 1, r);
            int lSum = Math.max(lStatus.lSum, lStatus.iSum + rStatus.lSum);
            int rSum = Math.max(rStatus.rSum, rStatus.iSum + lStatus.rSum);
            int mSum = Math.max(
                    // 不跨区比较
                    Math.max(lStatus.mSum, rStatus.mSum),
                    // 跨区比较
                    Math.max(
                            Math.max(lSum, rSum),
                            lStatus.rSum + rStatus.lSum
                    )
            );
            int iSum = lStatus.iSum + rStatus.iSum;
            return new Status(lSum, rSum, mSum, iSum);
        }

    }

    /**
     * [l,r]区间状态
     */
    class Status {
        // 表示 [l,r] 内以 l 为左端点的最大子段和
        private final int lSum;
        // 表示 [l,r] 内以 r 为右端点的最大子段和
        private final int rSum;
        // 表示 [l,r] 内的最大子段和
        private final int mSum;
        // 表示 [l,r] 的区间和
        private final int iSum;

        public Status(int lSum, int rSum, int mSum, int iSum) {
            this.lSum = lSum;
            this.rSum = rSum;
            this.mSum = mSum;
            this.iSum = iSum;
        }
    }


}