leetcode|152. Maximum Product Subarray[动态规划]-优快云博客

本文探讨了寻找具有最大乘积的连续子数组的问题，通过动态规划方法解决，避免了记录所有子数组值的内存限制问题，实现了空间复杂度为O(1)的高效算法。

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Problem

Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product.

Example 1:

Input: [2,3,-2,4]
Output: 6
Explanation: [2,3] has the largest product 6.
Example 2:

Input: [-2,0,-1]
Output: 0
Explanation: The result cannot be 2, because [-2,-1] is not a subarray.

Analysis

Dp - 记录局部最优子结构 - 动态状态转移，集greedy和recursive大成
想好dp状态的含义：

（Memory Limit Exceeded）第一反应，记录所有subarray的值，然后比较；dp[i,j]表示nums[i] * nums[i+1] … * nums[j] 的? dp[i,j] = dp[i,j-1]*dp[j,j]

// dp[i,j] = dp[i,j-1] * dp[j,j]
    public int maxProduct(int[] nums) {
        int len = nums.length;
        if (len < 1) return 0;
        int[][] dp = new int[len][len];
        for (int i = 0; i < len; i++) {
            for (int j = 0; j < len; j++) {
                if (j < i) dp[i][j] = 1;
                else if (j == i) dp[i][j] = nums[j];
                else dp[i][j] = dp[i][j-1]*nums[j];
            }
        }
        int res=Integer.MIN_VALUE;
        for(int i=0;i < len; i++){
            for (int j = 0; j < len; j++) {
                if(j<i) continue;
                res = Math.max(res, dp[i][j]);
            }
        }
        return res;
    }

dpMax[i]表示[k, i]子数组最大值，1<= k <= i
dpMin[i]表示[k, i]子数组最小值，1<= k <= i
dpMax=Math.max(Math.max(dpMax* nums[i], dpMin* nums[i]), nums[i]);
dpMin= Math.min(Math.min(dpMax * nums[i], dpMin * nums[i]), nums[i]);

空间复杂度进一步降低，因为只需要两个状态最大和最小，之前的状态都不需要保留。空间复杂度O(1)

public int maxProduct(int[] nums) {
        int len = nums.length;
        if (len < 1) return 0;
//        int[] dpMax = new int[len];
//        int[] dpMin = new int[len];
        int dpMax = nums[0];
        int dpMin = nums[0];
        int res = dpMax;
        for (int i = 1; i < len; i++) {
            int tmp = Math.max(Math.max(dpMax* nums[i], dpMin* nums[i]), nums[i]);
            dpMin= Math.min(Math.min(dpMax * nums[i], dpMin * nums[i]), nums[i]);
            dpMax = tmp;
            res = Math.max(res, dpMax);
        }
        return res;
    }