顺序遍历，数组求和，求解《927. 三等分》
例题
927. 三等分
给定一个由 0 和 1 组成的数组 arr ，将数组分成 3 个非空的部分 ，使得所有这些部分表示相同的二进制值。
如果可以做到，请返回任何 [i, j]，其中 i+1 < j，这样一来：
arr[0], arr[1], ..., arr[i] 为第一部分；
arr[i + 1], arr[i + 2], ..., arr[j - 1] 为第二部分；
arr[j], arr[j + 1], ..., arr[arr.length - 1] 为第三部分。
这三个部分所表示的二进制值相等。
如果无法做到，就返回 [-1, -1]。
注意，在考虑每个部分所表示的二进制时，应当将其看作一个整体。例如，[1,1,0] 表示十进制中的 6，而不会是 3。此外，前导零也是被允许的，所以 [0,1,1] 和 [1,1] 表示相同的值。
示例 1：
输入：arr = [1,0,1,0,1]
输出：[0,3]
示例 2：
输入：arr = [1,1,0,1,1]
输出：[-1,-1]
示例 3:
输入：arr = [1,1,0,0,1]
输出：[0,2]
提示：
3 <= arr.length <= 3 * 104
arr[i] 是 0 或 1
答案
顺序遍历
var threeEqualParts = function(arr) {
const n = arr.length, sum = _.sum(arr)
if (sum === 0) return [0, 2]
if (sum % 3 !== 0) return [-1, -1]
const avg = _.sum(arr) / 3
let first = 0, second = 0, third = 0
for (let i = 0, sum = 0; i < n; i++) {
if (arr[i]) {
if (sum === 0) first = i
else if (sum === avg) second = i
else if (sum === avg << 1) {
third = i
break
}
sum += arr[i]
}
}
const len = n - third
if (first + len > second || second + len > third) return [-1, -1]
for (let i = 1; i < len; i++) {
if (arr[first + i] !== arr[second + i] || arr[second + i] !== arr[third + i]) return [-1, -1]
}
return [first + len - 1, second + len]
};var threeEqualParts = function(arr) {
const n = arr.length
let sum = 0
for (let i = 0; i < n; i++) sum += arr[i]
if (sum === 0) return [0, 2]
if (sum % 3 !== 0) return [-1, -1]
const avg = sum / 3
let first = 0, second = 0, third = 0
for (let i = 0, sum = 0; i < n; i++) {
if (arr[i]) {
if (sum === 0) first = i
else if (sum === avg) second = i
else if (sum === avg << 1) {
third = i
break
}
sum += arr[i]
}
}
const len = n - third
if (first + len > second || second + len > third) return [-1, -1]
for (let i = 1; i < len; i++) {
if (arr[first + i] !== arr[second + i] || arr[second + i] !== arr[third + i]) return [-1, -1]
}
return [first + len - 1, second + len]
};function threeEqualParts(arr: number[]): number[] {
const n = arr.length
let sum = 0
for (let i = 0; i < n; i++) sum += arr[i]
if (sum === 0) return [0, 2]
if (sum % 3) return [-1, -1]
const avg = sum / 3
let first = 0, second = 0, third = 0
for (let i = 0, sum = 0; i < n; i++) {
if (arr[i] === 0) continue
if (sum === 0) first = i
else if (sum === avg) second = i
else if (sum === avg << 1) {
third = i
break
}
sum += arr[i]
}
const len = n - third
if (first + len > second || second + len > third) return [-1, -1]
for (let i = 1; i < len; i++) {
if (arr[first + i] !== arr[second + i] || arr[second + i] !== arr[third + i]) return [-1, -1]
}
return [first + len - 1, second + len]
};class Solution {
function threeEqualParts($arr) {
$sum = array_sum($arr);
if ($sum === 0) return [0, 2];
if ($sum % 3) return [-1, -1];
$avg = $sum / 3;
$total = 0;
$first = $second = $third = 0;
foreach ($arr as $i => $num) {
if ($num === 0) continue;
if ($total === 0) $first = $i;
elseif ($total === $avg) $second = $i;
elseif ($total === $avg << 1) {
$third = $i;
break;
}
$total += $num;
}
$len = count($arr) - $third;
if ($first + $len > $second || $second + $len > $third) return [-1, -1];
for ($i = 1; $i < $len; $i++) {
if ($arr[$first + $i] !== $arr[$second + $i] || $arr[$second + $i] !== $arr[$third + $i]) return [-1, -1];
}
return [$first + $len - 1, $second + $len];
}
}func threeEqualParts(arr []int) []int {
sum := 0
for _, num := range arr {
sum += num
}
if sum == 0 {
return []int{0, 2}
}
if sum % 3 != 0 {
return []int{-1, -1}
}
first, second, third, total, avg := 0, 0, 0, 0, sum / 3
label: for i, num := range arr {
if num == 0 {
continue
}
switch total {
case 0:
first = i
case avg:
second = i
case avg << 1:
third = i
break label
}
total += num
}
size := len(arr) - third
if first + size > second || second + size > third {
return []int{-1, -1}
}
for i := 1; i < size; i++ {
if arr[first + i] != arr[second + i] || arr[second + i] != arr[third + i] {
return []int{-1, -1}
}
}
return []int{first + size - 1, second + size}
}class Solution {
public int[] threeEqualParts(int[] arr) {
int sum = Arrays.stream(arr).sum();
if (sum == 0) return new int[]{0, 2};
if (sum % 3 > 0) return new int[]{-1, -1};
int avg = sum / 3, first = 0, second = 0, third = 0, total = 0;
int n = arr.length;
for (int i = 0; i < n; i++) {
if (arr[i] == 0) continue;
if (total == 0) first = i;
else if (total == avg) second = i;
else if (total == avg << 1) {
third = i;
break;
}
total += arr[i];
}
int len = n - third;
if (first + len > second || second + len > third) return new int[]{-1, -1};
for (int i = 1; i < len; i++) {
if (arr[first + i] != arr[second + i] || arr[second + i] != arr[third + i]) return new int[]{-1, -1};
}
return new int[]{first + len - 1, second + len};
}
}public class Solution {
public int[] ThreeEqualParts(int[] arr) {
int sum = arr.Sum();
if (sum == 0) return new int[]{0, 2};
if (sum % 3 > 0) return new int[]{-1, -1};
int avg = sum / 3, first = 0, second = 0, third = 0, total = 0, n = arr.Length;
for (int i = 0; i < n; i++) {
if (arr[i] == 0) continue;
if (total == 0) first = i;
else if (total == avg) second = i;
else if (total == avg << 1) {
third = i;
break;
}
total += 1;
}
int len = n - third;
if (first + len > second || second + len > third) return new int[]{-1, -1};
for (int i = 1; i < len; i++) {
if (arr[first + i] != arr[second + i] || arr[second + i] != arr[third + i]) return new int[]{-1, -1};
}
return new int[]{first + len - 1, second + len};
}
}int* threeEqualParts(int* arr, int arrSize, int* returnSize){
int sum = 0;
int* r = malloc(sizeof(int) * 2);
*returnSize = 2;
memset(r, -1, sizeof(int) * 2);
for (int i = 0; i < arrSize; i++) sum += arr[i];
if (sum == 0) {
r[0] = 0;
r[1] = 2;
return r;
}
if (sum % 3) return r;
int avg = sum / 3, first = 0, second = 0, third = 0, total = 0;
for (int i = 0; i < arrSize; i++) {
if (arr[i] == 0) continue;
if (total == 0) first = i;
else if (total == avg) second = i;
else if (total == avg << 1) {
third = i;
break;
}
total += 1;
}
int len = arrSize - third;
if (first + len > second || second + len > third) return r;
for (int i = 1; i < len; i++) {
if (arr[first + i] != arr[second + i] || arr[second + i] != arr[third + i]) return r;
}
r[0] = first + len - 1;
r[1] = second + len;
return r;
}class Solution {
public:
vector<int> threeEqualParts(vector<int>& arr) {
int sum = accumulate(arr.begin(), arr.end(), 0);
if (sum == 0) return {0, 2};
if (sum % 3) return {-1, -1};
int avg = sum / 3, first = 0, second = 0, third = 0, total = 0, n = arr.size();
for (int i = 0; i < n; i++) {
if (arr[i] == 0) continue;
if (total == 0) first = i;
else if (total == avg) second = i;
else if (total == avg << 1) {
third = i;
break;
}
total += 1;
}
int len = n - third;
if (first + len > second || second + len > third) return {-1, -1};
for (int i = 1; i < len; i++) {
if (arr[first + i] != arr[second + i] || arr[second + i] != arr[third + i]) return {-1, -1};
}
return {first + len - 1, second + len};
}
};class Solution:
def threeEqualParts(self, arr: List[int]) -> List[int]:
summation = sum(arr)
if summation == 0: return [0, 2]
if summation % 3: return [-1, -1]
avg, first, second, third, total = summation / 3, 0, 0, 0, 0
for i, num in enumerate(arr):
if num == 0: continue
if total == 0: first = i
elif total == avg: second = i
elif total == int(avg) << 1:
third = i
break
total += 1
size = len(arr) - third
if first + size > second or second + size > third: return [-1, -1]
for i in range(1, size):
if arr[first + i] != arr[second + i] or arr[second + i] != arr[third + i]: return [-1, -1]
return [first + size - 1, second + size]