[LeetCode]53. Maximum Subarray

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本文介绍了使用贪心算法和动态规划求解最大连续子数组和的问题，包括两种不同的实现方式，并提供了代码示例。

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.

贪心算法1：

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

贪心算法2：

 public int maxSubArray(int[] nums) {
     int sum = nums[0];
     int max = nums[0];
     for (int i = 1; i< nums.length; i++) {
         sum = sum < 0 ? nums[i] : sum + nums[i];
         max = max > sum ? max : sum;
     }
     return max;
}

动态规划：

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