1 暴力解

时间复杂度：O(n3)

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

2 改进的暴力解

时间复杂度：O(n2)

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

3 分治

时间复杂度：O(nlog⁡n)

4 贪心

时间复杂度：O(n)

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

5 动态规划

时间复杂度：O(n)

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class Solution {
    public int maxSubArray(int[] nums) {
        int max = Integer.MIN_VALUE;
        int[] dp = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            if (i == 0 || dp[i - 1] < 0) {
                dp[i] = nums[i];
            } else {
                dp[i] = nums[i] + dp[i - 1];
            }
            max = Math.max(max, dp[i]);
        }
        return max;
    }
}

附录

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