525. Contiguous Array
Given a binary array nums
, return the maximum length of a contiguous subarray with an equal number of 0
and1
.
Example 1:
Input: nums = [0,1] Output: 2 Explanation: [0, 1] is the longest contiguous subarray with an equal number of 0 and 1.
Example 2:
Input: nums = [0,1,0] Output: 2 Explanation: [0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.
Constraints:
1 <= nums.length <= 105
nums[i]
is either0
or1
.
class Solution {
public:
int findMaxLength(vector<int>& nums) {
int res = 0;
int n = nums.size();
int sum = 0;
unordered_map<int, int> m{{0, -1}};
for (int i = 0; i < n; ++i) {
// 遇到 1 就 +1,遇到 0 就 -1
sum += (nums[i] == 1) ? 1 : -1;
// 如果加起來的總和不等於 0 的話
// 目前的長度為目前的 index i 減掉 m[sum]
if (m.count(sum)) {
res = max(res, i - m[sum]);
} else {
// 如果加起來的總和等於 0 的話,其值為目前的 index
m[sum] = i;
}
}
// nums = [0,1,0]
// m = {{-1: 0}, {0: -1}}
// nums = [0, 0, 0, 1, 1, 1]
// m = {{-2: 1}, {-1: 0}, {-3: 2}, {0: -1}}
return res;
}
};
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