[Lintcode]Maximum Subarray

Given an array of integers, find a contiguous subarray which has the largest sum

Example

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

Note

题目要求O(n)复杂度.

滑动窗口法。如果sum < 0, 那么令sum = 0重新开始。因为以负数为基数的话，无论如何加减，都比以0为基数小。但是仍然将负数与max取最大值并保存，用来防止全部为负数的情况。

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

另外还可以用一维DP解决，只是空间复杂度为O（n）。注意数组最后的值未必是最大值

public class Solution {    /**     * @param nums: a list of integers     * @return: A integer indicate the sum of minimum subarray     */    public int maxSubArray(int[] nums) {        // write your code        if(nums.length == 0)              return 0;                      int n = nums.length;        int []global = new int[n];                global[0] = nums[0];        int res = global[0];        for(int i=1; i<n; i++)          {              global[i] = Math.max(nums[i], global[i - 1] + nums[i]);             res = Math.max(res, global[i]);        }          return res > global[n - 1]? res : global[n - 1];      }}

另外在网上看到一种算法, 原理与第一种算法类似。第一种算法为每次sum小于0则重新开始。这个算法为在sum最小值处重新开始。

public class Solution {    public int maxSubArray(int[] A) {        if (A == null || A.length == 0){            return 0;        }                int max = Integer.MIN_VALUE, sum = 0, minSum = 0;        for (int i = 0; i < A.length; i++) {            sum += A[i];            max = Math.max(max, sum - minSum);            minSum = Math.min(minSum, sum);        }        return max;    }}