LeetCode 53. Maximum Subarray（最大子数组）

原题网址：https://leetcode.com/problems/maximum-subarray/

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.

方法一：动态规划/贪心法/Kadane算法，如果是负数，则抛弃前面所有的数字之和，重新开始计算。

public class Solution {
public int maxSubArray(int[] nums) {
int sum = nums[0];
int t = sum;
for(int i=1; i<nums.length; i++) {
if (t < 0) t = 0;
t += nums[i];
if (t>sum) sum = t;
}
return sum;
}
}

方法二：分治策略。

public class Solution {
private Result max(int[] nums, int from, int to) {
Result result = new Result();
if (from==to) {
result.max = nums[from];
result.lmax = nums[from];
result.rmax = nums[from];
result.sum = nums[from];
return result;
}
int m = (from+to)/2;
Result r1 = max(nums, from, m);
Result r2 = max(nums, m+1, to);
result.max = Math.max(Math.max(r1.max, r2.max), r1.rmax+r2.lmax);
result.lmax = Math.max(r1.lmax, r1.sum+r2.lmax);
result.rmax = Math.max(r1.rmax+r2.sum, r2.rmax);
result.sum = r1.sum+r2.sum;
return result;
}
public int maxSubArray(int[] nums) {
if (nums == null || nums.length == 0) return 0;
Result result = max(nums, 0, nums.length-1);
return result.max;
}
}
class Result {
// 最大和
int max;
// 靠左侧最大和
int lmax;
// 靠右侧最大和
int rmax;
// 数组和
int sum;
}