将数组分为三个部分，每部分对应二进制数值相同
1. 三指针
各部分1的数目应当相同，借助这个性质来找三个指针的起始位置
根据最后一个指针起始位置确定串的长度，逐位比较即可
class Solution {
public:
vector<int> threeEqualParts(vector<int>& arr) {
int sum = accumulate(arr.begin(), arr.end(), 0);
if (sum % 3 != 0) return {-1, -1};//无法划分
if (sum == 0) return {0, 2}; //任意划分
int partial = sum / 3; //每个区域累计1的数目
int first = 0, second = 0, third = 0, cur = 0;
for (int i = 0; i < arr.size(); i++) { //去掉前导0
if (arr[i] == 1) { //累计1的数目
if (cur == 0) first = i;
else if (cur == partial) second = i;
else if (cur == 2 * partial) third = i;
cur++;
}
}
int len = (int)arr.size() - third;//第三个部分长度
if (first + len <= second && second + len <= third) {//去除前导0后，前面应该有足够的余量，最后三个串长度一致
int i = 0;
while (third + i < arr.size()) {
if (arr[first + i] != arr[second + i] || arr[first + i] != arr[third + i]) //同时比较对应位置
return {-1, -1};
i++;
}
return {first + len - 1, second + len};
}
return {-1, -1};
}
};