lintcode-45-最大子数组差

45-最大子数组差

给定一个整数数组，找出两个不重叠的子数组A和B，使两个子数组和的差的绝对值|SUM(A) - SUM(B)|最大。

返回这个最大的差值。

注意事项

子数组最少包含一个数

样例

给出数组[1, 2, -3, 1]，返回 6

挑战

时间复杂度为O(n)，空间复杂度为O(n)

标签

贪心 枚举法 LintCode 版权所有 数组 子数组 前后遍历

思路

复用“求最大字数组和”的代码,求出从左自右遍历数的最大子数组maxLeft、最小子数组minLeft和从右自左遍历数的最大子数组maxRight、最小子数组minRight。然后求出maxLeft与minRight的最大差，minLeft与maxRight的最大差。即可得出最大子数组差。

code

class Solution {
public:
    /**
     * @param nums: A list of integers
     * @return: An integer indicate the value of maximum difference between two
     *          Subarrays
     */
    int maxDiffSubArrays(vector<int> nums) {
        // write your code here
        int size = nums.size(), i = 0;
        int sum = 0, maxValue = 0;
        int dec = 0, minValue = 0;

        int *maxLeft = new int[size];
        int *minLeft = new int[size];
        int *maxRight = new int[size];
        int *minRight = new int[size];

        sum = maxValue = maxLeft[0] = nums[0];
        for(i=1; i<size; i++) {
            if(sum < 0) {
                sum = nums[i];
            }
            else {
                sum += nums[i];
            }
            if(sum > maxValue) {
                maxValue = sum;
            }
            maxLeft[i] = maxValue;
        }
        sum = maxValue = maxRight[size-1] = nums[size-1];
        for(i=size-2; i>=0; i--) {
            if(sum < 0) {
                sum = nums[i];
            }
            else {
                sum += nums[i];
            }
            if(sum > maxValue) {
                maxValue = sum;
            }
            maxRight[i] = maxValue;
        }

        dec = minValue = minLeft[0] = nums[0];
        for(i=1; i<size; i++) {
            if(dec > 0) {
                dec = nums[i];
            }
            else {
                dec += nums[i];
            }
            if(dec < minValue) {
                minValue = dec;
            }
            minLeft[i] = minValue;
        }
        dec = minValue = minRight[size-1] = nums[size-1];
        for(i=size-2; i>=0; i--) {
            if(dec > 0) {
                dec = nums[i];
            }
            else {
                dec += nums[i];
            }
            if(dec < minValue) {
                minValue = dec;
            }
            minRight[i] = minValue;
        }

        int result1 = 0x80000000;
        int result2 = 0x80000000;
        for(i=0; i<size-1; i++) {
            result1 = (result1 > maxLeft[i]-minRight[i+1])?result1:( maxLeft[i]-minRight[i+1]);
            result2 = (result2 > maxRight[i+1]-minLeft[i])?result2:( maxRight[i+1]-minLeft[i]);
        }

        delete[] maxLeft;
        delete[] minLeft;
        delete[] maxRight;
        delete[] minRight;

        return (result1 > result2)?result1:result2;
    }
};