LeetCode 53. Maximum Subarray--Divide and Conquer(分治法)

题目链接

53. 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.

解：最简单的思路是遍历每一段的和，时间复杂度是O(n^2)，优化了之后还是不能通过最后一个测试样例，超时，于是只能换一种思路了：

1.首先从左到右遍历数组，并记录一个[0,x]的和，如果计算的和相比之前的和大，那么就用大的和替换之前的和，并记录当前的x值（tari)，遍历结束时，所得到的tari的意义就是所求最大子数组如果是[0,tari]的子数组，其右边界一定是tari

2.从右到左遍历数组，同理得到一个tarj

3.如果tari >= tarj，易知所求子数组一定是[tarj, tari]

4.若tari < tarj，所求子数组一定在[0,tari],[tari,tarj],[tarj,size-1]中，再分别用同样的方法求得三个最大和，取其中最大值

代码：

class Solution {public:    int maxSubArray(vector<int>& nums) {        int i, j;        int tari, tarj;        int size = nums.size();        int sum = -2147483648;        int temp = 0;        for (i = 0; i < size; i++) {            temp += nums[i];            if (temp > sum) {                sum = temp;                tari = i;            }        }        temp = 0;        sum = -2147483648;        for (i = size-1; i >= 0; i--) {            temp += nums[i];            if (temp > sum) {                sum = temp;                tarj = i;            }        }        if (tari >= tarj) {        int result = 0;        for (j = tarj; j <= tari; j++) {                result += nums[j];            }            return result;        }        temp = 0;        int tarii;        sum = -2147483648;        for (j = tari; j >=0; j--) {            temp += nums[j];            if (temp > sum) {                sum = temp;                tarii = j;            }        }    int result1 = 0;        for (j = tarii; j <= tari; j++) result1 += nums[j];        temp = 0;        int tarjj;        sum = -2147483648;        for (j = tarj; j < size; j++) {            temp += nums[j];            if (temp > sum) {                sum = temp;                tarjj = j;            }        }        int result2 = 0;        for (j = tarj; j <= tarjj; j++) result2 += nums[j];        vector<int> nums1(nums.begin()+tari,nums.begin()+tarj);          int result3 = maxSubArray(nums1);        return max(max(result1, result2), result3);    }};