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Sep 25, 2025
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feat: update solutions to lc problem: No.0611 #4748

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141 changes: 57 additions & 84 deletions solution/0600-0699/0611.Valid Triangle Number/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -29,7 +29,7 @@ tags:
<pre>
<strong>输入:</strong> nums = [2,2,3,4]
<strong>输出:</strong> 3
<strong>解释:</strong>有效的组合是:
<strong>解释:</strong>有效的组合是:
2,3,4 (使用第一个 2)
2,3,4 (使用第二个 2)
2,2,3
Expand Down Expand Up @@ -58,13 +58,19 @@ tags:

### 方法一：排序 + 二分查找

一个有效三角形需要满足：**任意两边之和大于第三边**。即：`a + b > c`①, `a + c > b`②, `b + c > a`③。
一个有效三角形需要满足：**任意两边之和大于第三边**。即：

如果我们将边按从小到大顺序排列，即 `a < b < c`，那么显然 ②③ 成立，我们只需要确保 ① 也成立，就可以形成一个有效三角形。
$$a + b \gt c \tag{1}$$

我们在 `[0, n - 3]` 范围内枚举 i，在 `[i + 1, n - 2]` 范围内枚举 j，在 `[j + 1, n - 1]` 范围内进行二分查找，找出第一个大于等于 `nums[i] + nums[j]` 的下标 left，那么在 `[j + 1, left - 1]` 范围内的 k 满足条件，将其累加到结果 ans。
$$a + c \gt b \tag{2}$$

时间复杂度：$O(n^2\log n)$。
$$b + c \gt a \tag{3}$$

如果我们将边按从小到大顺序排列，即 $a \leq b \leq c$，那么显然 (2)(3) 成立，我们只需要确保 (1) 也成立，就可以形成一个有效三角形。

我们在 $[0, n - 3]$ 范围内枚举 i，在 $[i + 1, n - 2]$ 范围内枚举 j，在 $[j + 1, n - 1]$ 范围内进行二分查找，找出第一个大于等于 $nums[i] + nums[j]$ 的下标 left，那么在 $[j + 1, left - 1]$ 范围内的 k 满足条件，将其累加到结果 $\textit{ans}$。

时间复杂度 $O(n^2\log n)$，空间复杂度 $O(\log n)$。其中 $n$ 是数组的长度。

<!-- tabs:start -->

Expand All @@ -88,20 +94,22 @@ class Solution:
class Solution {
public int triangleNumber(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
int res = 0;
for (int i = n - 1; i >= 2; --i) {
int l = 0, r = i - 1;
while (l < r) {
if (nums[l] + nums[r] > nums[i]) {
res += r - l;
--r;
} else {
++l;
int ans = 0;
for (int i = 0, n = nums.length; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
int left = j + 1, right = n;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[mid] >= nums[i] + nums[j]) {
right = mid;
} else {
left = mid + 1;
}
}
ans += left - j - 1;
}
}
return res;
return ans;
}
}
```
Expand All @@ -112,12 +120,14 @@ class Solution {
class Solution {
public:
int triangleNumber(vector<int>& nums) {
sort(nums.begin(), nums.end());
ranges::sort(nums);
int ans = 0, n = nums.size();
for (int i = 0; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
int k = lower_bound(nums.begin() + j + 1, nums.end(), nums[i] + nums[j]) - nums.begin() - 1;
ans += k - j;
int sum = nums[i] + nums[j];
auto it = ranges::lower_bound(nums.begin() + j + 1, nums.end(), sum);
int k = int(it - nums.begin()) - 1;
ans += max(0, k - j);
}
}
return ans;
Expand All @@ -130,19 +140,15 @@ public:
```go
func triangleNumber(nums []int) int {
sort.Ints(nums)
n := len(nums)
ans := 0
for i, n := 0, len(nums); i < n-2; i++ {
for i := 0; i < n-2; i++ {
for j := i + 1; j < n-1; j++ {
left, right := j+1, n
for left < right {
mid := (left + right) >> 1
if nums[mid] >= nums[i]+nums[j] {
right = mid
} else {
left = mid + 1
}
sum := nums[i] + nums[j]
k := sort.SearchInts(nums[j+1:], sum) + j + 1 - 1
if k > j {
ans += k - j
}
ans += left - j - 1
}
}
return ans
Expand All @@ -154,17 +160,14 @@ func triangleNumber(nums []int) int {
```ts
function triangleNumber(nums: number[]): number {
nums.sort((a, b) => a - b);
let n = nums.length;
const n = nums.length;
let ans = 0;
for (let i = n - 1; i >= 2; i--) {
let left = 0,
right = i - 1;
while (left < right) {
if (nums[left] + nums[right] > nums[i]) {
ans += right - left;
right--;
} else {
left++;
for (let i = 0; i < n - 2; i++) {
for (let j = i + 1; j < n - 1; j++) {
const sum = nums[i] + nums[j];
let k = _.sortedIndex(nums, sum, j + 1) - 1;
if (k > j) {
ans += k - j;
}
}
}
Expand All @@ -179,56 +182,26 @@ impl Solution {
pub fn triangle_number(mut nums: Vec<i32>) -> i32 {
nums.sort();
let n = nums.len();
let mut res = 0;
for i in (2..n).rev() {
let mut left = 0;
let mut right = i - 1;
while left < right {
if nums[left] + nums[right] > nums[i] {
res += right - left;
right -= 1;
} else {
left += 1;
}
}
}
res as i32
}
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### 方法二

<!-- tabs:start -->

#### Java

```java
class Solution {
public int triangleNumber(int[] nums) {
Arrays.sort(nums);
int ans = 0;
for (int i = 0, n = nums.length; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
int left = j + 1, right = n;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[mid] >= nums[i] + nums[j]) {
right = mid;
} else {
let mut ans = 0;
for i in 0..n.saturating_sub(2) {
for j in i + 1..n.saturating_sub(1) {
let sum = nums[i] + nums[j];
let mut left = j + 1;
let mut right = n;
while left < right {
let mid = (left + right) / 2;
if nums[mid] < sum {
left = mid + 1;
} else {
right = mid;
}
}
ans += left - j - 1;
if left > j + 1 {
ans += (left - 1 - j) as i32;
}
}
}
return ans;
ans
}
}
```
Expand Down
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