LintCode 994: Contiguous Array 前缀和数组典型题-优快云博客

本文介绍了一种算法，用于在给定的二进制数组中找到最长具有相同数量0和1的连续子数组。使用presums数组和hashmap来存储累积和及其出现的位置，以找到最大长度。

994 · Contiguous Array
Algorithms
Medium

Description
Given a binary array, find the maximum length of a contiguous subarray with equal number of 0 and 1.
The length of the given binary array will not exceed 50,000.

Example
Example 1:

Input: [0,1]
Output: 2
Explanation: [0, 1] is the longest contiguous subarray with equal number of 0 and 1.
Example 2:

Input: [0,1,0]
Output: 2
Explanation: [0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.

解法1：用presums数组。注意可以把0变成-1, 并利用hashmap，这样只要有两个presum是一样的就肯定是相同的0和1。注意hashmap里面存的对应的index是第一次出现的presum的index。

class Solution {
public:
    /**
     * @param nums: a binary array
     * @return: the maximum length of a contiguous subarray
     */
    int findMaxLength(vector<int> &nums) {
       int n = nums.size();
       vector<int> presums(n + 1, 0);
       map<int, int> mp; //presums, index
       int res = 0;
       mp[0] = 0;
       for (int i = 1; i <= n; i++) {
           presums[i] = presums[i - 1] + (nums[i - 1] == 0 ? -1 : 1);
           if (mp.find(presums[i]) == mp.end()) {
               mp[presums[i]] = i;
           } else {
              res = max(res, i - mp[presums[i]]);
           }
       }
       return res;
    }
};