leetcode 525. Contiguous Array

Given a binary array, find the maximum length of a contiguous subarray with equal number of 0 and 1.

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.

Note: The length of the given binary array will not exceed 50,000.

这道题我想用DP做的，结果Memory Limit Exceeded了，真是屮艸芔茻...

注意一个测试用例：[0,0,0,1,1,1,0]，返回结果 6 。

这道题有点类似于之前的一道题 http://blog.csdn.net/huanghanqian/article/details/77054149，从树的根一直遍历到叶，找到所有的 和为target的路径(不一定非得以根结点开始以叶节点结尾）。只要把这道题的0想象成-1，就会转化为：找到最长的串使得 和为0。然后解决方法就跟之前的那道题类似啦。

The idea is to change 0 in the original array to -1. Thus, if we find SUM[i, j] == 0 then we know there are even number of -1 and 1 between index i and j. Also put the sum to index mapping to a HashMap to make search faster.

public int findMaxLength(int[] nums) {
	int n=nums.length;
	if(n<2){
		return 0;
	}
	int maxLength=0;
	//map存储 <[0,index]范围内的和,index> ，且 是 得到该和的最小的index
	HashMap map=new HashMap();
	int sum=0;//sum是0~i范围的和
	map.put(0, -1);
	for(int i=0;imaxLength){
				maxLength=length;
			}
		}
		else {
			map.put(sum,i);
		}
	}
	return maxLength;
}

大神们也都用的这个解法。

这道题有solutions： https://leetcode.com/problems/contiguous-array/solution/#approach-3-using-hashmap-accepted

Approach #3 Using HashMap [Accepted]

Algorithm

我们使用一个 HashMap $map$ 来存储 $(index,count)，$当我们遇到一个之前未曾出现的count时，将该条目push进map。之后再碰到该count时，使用它对应的 index 来找 最大的使得count为0的子串。

Java

public class Solution {

    public int findMaxLength(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, -1);
        int maxlen = 0, count = 0;
        for (int i = 0; i < nums.length; i++) {
            count = count + (nums[i] == 1 ? 1 : -1);
            if (map.containsKey(count)) {
                maxlen = Math.max(maxlen, i - map.get(count));
            } else {
                map.put(count, i);
            }
        }
        return maxlen;
    }
}

Complexity Analysis

• Time complexity : $O(n)$. The entire array is traversed only once.

• Space complexity : $O(n)$. Maximum size of the HashMap $map$ will be $\text{n}$, if all the elements are either 1 or 0.