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37 | 37 |
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38 | 38 | ## Solutions
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39 | 39 |
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| 40 | +similar problem: [LIS](../../0300-0399/0300.Longest%20Increasing%20Subsequence/README_EN.md) |
| 41 | + |
| 42 | +`cnt` array records the number of longest sequence ending in `nums[i]` |
| 43 | + |
40 | 44 | <!-- tabs:start -->
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41 | 45 |
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42 | 46 | ### **Python3**
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43 | 47 |
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44 | 48 | ```python
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45 |
| - |
| 49 | +class Solution: |
| 50 | + def findNumberOfLIS(self, nums: List[int]) -> int: |
| 51 | + maxLen, ans, n = 0, 0, len(nums) |
| 52 | + dp, cnt = [1] * n, [1] * n |
| 53 | + for i in range(n): |
| 54 | + for j in range(i): |
| 55 | + if nums[i] > nums[j]: |
| 56 | + if dp[j] + 1 > dp[i]: |
| 57 | + dp[i] = dp[j] + 1 |
| 58 | + cnt[i] = cnt[j] |
| 59 | + elif dp[j] + 1 == dp[i]: |
| 60 | + cnt[i] += cnt[j] |
| 61 | + if dp[i] > maxLen: |
| 62 | + maxLen = dp[i] |
| 63 | + ans = cnt[i] |
| 64 | + elif dp[i] == maxLen: |
| 65 | + ans += cnt[i] |
| 66 | + return ans |
46 | 67 | ```
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47 | 68 |
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48 | 69 | ### **Java**
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49 | 70 |
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50 | 71 | ```java
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| 72 | +class Solution { |
| 73 | + public int findNumberOfLIS(int[] nums) { |
| 74 | + int maxLen = 0, ans = 0, n = nums.length; |
| 75 | + int[] dp = new int[n]; |
| 76 | + int[] cnt = new int[n]; |
| 77 | + for (int i = 0; i < n; i++) { |
| 78 | + dp[i] = 1; |
| 79 | + cnt[i] = 1; |
| 80 | + for (int j = 0; j < i; j++) { |
| 81 | + if (nums[i] > nums[j]) { |
| 82 | + if (dp[j] + 1 > dp[i]) { |
| 83 | + dp[i] = dp[j] + 1; |
| 84 | + cnt[i] = cnt[j]; |
| 85 | + } else if (dp[j] + 1 == dp[i]) { |
| 86 | + cnt[i] += cnt[j]; |
| 87 | + } |
| 88 | + } |
| 89 | + } |
| 90 | + if (dp[i] > maxLen) { |
| 91 | + maxLen = dp[i]; |
| 92 | + ans = cnt[i]; |
| 93 | + } else if (dp[i] == maxLen) { |
| 94 | + ans += cnt[i]; |
| 95 | + } |
| 96 | + } |
| 97 | + return ans; |
| 98 | + } |
| 99 | +} |
| 100 | +``` |
| 101 | + |
| 102 | +### **C++** |
| 103 | + |
| 104 | +```cpp |
| 105 | +class Solution { |
| 106 | +public: |
| 107 | + int findNumberOfLIS(vector<int>& nums) { |
| 108 | + int maxLen = 0, ans = 0, n = nums.size(); |
| 109 | + vector<int> dp(n, 1), cnt(n, 1); |
| 110 | + for (int i = 0; i < n; ++i) { |
| 111 | + for (int j = 0; j < i; ++j) { |
| 112 | + if (nums[i] > nums[j]) { |
| 113 | + if (dp[j] + 1 > dp[i]) { |
| 114 | + dp[i] = dp[j] + 1; |
| 115 | + cnt[i] = cnt[j]; |
| 116 | + } else if (dp[j] + 1 == dp[i]) { |
| 117 | + cnt[i] += cnt[j]; |
| 118 | + } |
| 119 | + } |
| 120 | + } |
| 121 | + if (dp[i] > maxLen) { |
| 122 | + maxLen = dp[i]; |
| 123 | + ans = cnt[i]; |
| 124 | + } else if (dp[i] == maxLen) { |
| 125 | + ans += cnt[i]; |
| 126 | + } |
| 127 | + } |
| 128 | + return ans; |
| 129 | + } |
| 130 | +}; |
| 131 | +``` |
| 132 | +
|
| 133 | +### **Go** |
| 134 | +
|
| 135 | +```go |
| 136 | +func findNumberOfLIS(nums []int) int { |
| 137 | + maxLen, ans, n := 0, 0, len(nums) |
| 138 | + dp, cnt := make([]int, n), make([]int, n) |
| 139 | + for i := 0; i < n; i++ { |
| 140 | + dp[i] = 1 |
| 141 | + cnt[i] = 1 |
| 142 | + for j := 0; j < i; j++ { |
| 143 | + if nums[i] > nums[j] { |
| 144 | + if dp[j]+1 > dp[i] { |
| 145 | + dp[i] = dp[j] + 1 |
| 146 | + cnt[i] = cnt[j] |
| 147 | + } else if dp[j]+1 == dp[i] { |
| 148 | + cnt[i] += cnt[j] |
| 149 | + } |
| 150 | + } |
| 151 | + } |
| 152 | + if dp[i] > maxLen { |
| 153 | + maxLen = dp[i] |
| 154 | + ans = cnt[i] |
| 155 | + } else if dp[i] == maxLen { |
| 156 | + ans += cnt[i] |
| 157 | + } |
| 158 | + } |
| 159 | + return ans |
| 160 | +} |
| 161 | +``` |
51 | 162 |
|
| 163 | +### **Rust** |
| 164 | + |
| 165 | +```rust |
| 166 | +impl Solution { |
| 167 | + pub fn find_number_of_lis(nums: Vec<i32>) -> i32 { |
| 168 | + let mut max_len = 0; |
| 169 | + let mut ans = 0; |
| 170 | + let n = nums.len(); |
| 171 | + let mut dp = vec![1; n]; |
| 172 | + let mut cnt = vec![1; n]; |
| 173 | + for i in 0..n { |
| 174 | + for j in 0..i { |
| 175 | + if nums[i] > nums[j] { |
| 176 | + if dp[j] + 1 > dp[i] { |
| 177 | + dp[i] = dp[j] + 1; |
| 178 | + cnt[i] = cnt[j]; |
| 179 | + } else if dp[j] + 1 == dp[i] { |
| 180 | + cnt[i] += cnt[j]; |
| 181 | + } |
| 182 | + } |
| 183 | + } |
| 184 | + if dp[i] > max_len { |
| 185 | + max_len = dp[i]; |
| 186 | + ans = cnt[i]; |
| 187 | + } else if dp[i] == max_len { |
| 188 | + ans += cnt[i]; |
| 189 | + } |
| 190 | + } |
| 191 | + ans |
| 192 | + } |
| 193 | +} |
52 | 194 | ```
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53 | 195 |
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54 | 196 | ### **...**
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