LeetCode——Maximum Subarray

Description：

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

click to show more practice.

More practice:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

看见这题的第一想法就是决不能用暴力穷举。

然后应该会想到简单的dp。思路就是每次都选择最大的sum（这不是贪心？）。时间复杂度是O(n)，空间复杂度是O(1)；

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

题目最后还有一个More practice:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

这样的话就要用分治和递归来解。

首先把问题分成若干子问题，找到子问题中的解后逐一合并直到得到整个问题的解。说起来挺简单但是细节问题还是要特别注意的。

时间复杂度是O(nlogn)，空间复杂度是O(logn)；

public class Solution {
    
    //分治
    public int divide(int[] nums, int m, int n) {
        
        if(m == n) {
            return nums[m];
        }
        
        int mid = m + (n-m)/2; //防止整数溢出
        
        int home = divide(nums, m, mid);
        int end = divide(nums, mid+1, n);
        
        int sum = merge(nums, m, n, mid);
        
        return max(sum, max(home, end));
        
    }
    
    //合并
    public int merge(int[] nums, int m, int n, int mid) {
        
        int leftMax = nums[mid];
        int sum = 0;
        for(int i=mid; i>=m; i--) {
            sum += nums[i];
            if(leftMax < sum) {
                leftMax = sum;
            }
        }
        
        sum = 0;
        
        int rightMax = nums[mid+1];
        for(int i=mid+1; i<=n; i++) {
            sum += nums[i];
            if(rightMax < sum) {
                rightMax = sum;
            }
        }
        
        sum = leftMax + rightMax;
        
        return sum;
    }
    
    public int max(int a, int b) {
        return a > b ? a : b;
    }
    
    public int maxSubArray(int[] nums) {
        
        if(nums.length <= 0) {
            return 0;
        }
        
        int res = divide(nums, 0, nums.length-1);
        
        return res;
    }
}

分治、dp、贪心有时候傻傻分不清楚。