LeetCode 53. Maximum Subarray

原题链接在这里：https://leetcode.com/problems/maximum-subarray/

题目：

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.

题解：

采用的是DP解法. 把原有问题拆成小问题. maxSubArray(int [] nums, int i). 表示到 [0,i]段最大的sub array. 前提是包括 i 元素.

maxSubArray(nums, i) = nums[i] + maxSubArray(nums, i-1) > 0 ? maxSubArray(nums, i-1) : 0.

同时维护全局最大值.

历史信息就是local[i-1] 和 global[i-1], 更新当前信息时先更新局部最优.

local[i] = Math.max(local[i-1]+nums[i], nums[i]), 有可能local[i-1]是负数，就直接取nums[i], 然后更新全局最优.

global[i] = Math.max(lcoal[i], global[i-1]), 要么是原来的全局最优，要么是局部最优，此时能涵盖所有解。如果全局最优不包含当前值，那么会被维护在global[i-1]中，若包含当前值，那么就是local[i].

Note: local, global的初始化是nums[0]. 不能初始化成0, 然后从i=1开始，比如[-1], Output:0, Expected:-1, 这种corner case 就会出错.

Time Complexity: O(n).n = nums.length.

Space: O(1).

AC Java:

public class Solution {
    public int maxSubArray(int[] nums) {
        if(nums == null || nums.length ==0){
            return 0;
        }
        int global = nums[0];
        int local = nums[0];
        for(int i = 1; i< nums.length; i++){
            local = Math.max(local + nums[i], nums[i]);
            global = Math.max(global, local);
        }
        return global;
    }
}

类似Maximum Product Subarray, Maximum Subarray Sum with One Deletion, House Robber, Best Time to Buy and Sell Stock., Longest Turbulent Subarray.

跟上Substring With Largest Variance.

跟上Maximum Subarray II, Max Sum of Rectangle No Larger Than K.

Reference: http://blog.csdn.net/linhuanmars/article/details/21314059