Merge pull request halfrost#71 from harryleonardo/master

halfrost · halfrost · commit c1f421273b76 · 2020-09-24T16:42:52.000+08:00
Add problem 1480 &amp; 1512
diff --git a/leetcode/1480.Running-Sum-of-1d-Array/1480. Running Sum of 1d Array.go b/leetcode/1480.Running-Sum-of-1d-Array/1480. Running Sum of 1d Array.go
@@ -0,0 +1,10 @@
+package leetcode
+
+func runningSum(nums []int) []int {
+	dp := make([]int, len(nums)+1)
+	dp[0] = 0
+	for i := 1; i <= len(nums); i++ {
+		dp[i] = dp[i-1] + nums[i-1]
+	}
+	return dp[1:]
+}
diff --git a/leetcode/1480.Running-Sum-of-1d-Array/1480. Running Sum of 1d Array_test.go b/leetcode/1480.Running-Sum-of-1d-Array/1480. Running Sum of 1d Array_test.go
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1480 struct {
+	para1480
+	ans1480
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1480 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1480 struct {
+	one []int
+}
+
+func Test_Problem1480(t *testing.T) {
+
+	qs := []question1480{
+
+		{
+			para1480{[]int{1, 2, 3, 4}},
+			ans1480{[]int{1, 3, 6, 10}},
+		},
+
+		{
+			para1480{[]int{1, 1, 1, 1, 1}},
+			ans1480{[]int{1, 2, 3, 4, 5}},
+		},
+
+		{
+			para1480{[]int{3, 1, 2, 10, 1}},
+			ans1480{[]int{3, 4, 6, 16, 17}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1480------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1480, q.para1480
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, runningSum(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1480.Running-Sum-of-1d-Array/README.md b/leetcode/1480.Running-Sum-of-1d-Array/README.md
@@ -0,0 +1,64 @@
+# [1480. Running Sum of 1d Array](https://leetcode.com/problems/running-sum-of-1d-array/)
+
+## 题目
+
+Given an array `nums`. We define a running sum of an array as `runningSum[i] = sum(nums[0]…nums[i])`.
+
+Return the running sum of `nums`.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,4]
+Output: [1,3,6,10]
+Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,1,1,1,1]
+Output: [1,2,3,4,5]
+Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [3,1,2,10,1]
+Output: [3,4,6,16,17]
+
+```
+
+
+**Constraints**:
+
+- `1 <= nums.length <= 1000`
+- `-10^6 <= nums[i] <= 10^6`
+
+## 题目大意
+
+给你一个数组 nums 。数组「动态和」的计算公式为：runningSum[i] = sum(nums[0]…nums[i]) 。请返回 nums 的动态和。
+
+
+## 解题思路
+
+- 简单题，按照题意依次循环计算前缀和即可。
+
+## 代码
+
+```go
+package leetcode
+
+func runningSum(nums []int) []int {
+	dp := make([]int, len(nums)+1)
+	dp[0] = 0
+	for i := 1; i <= len(nums); i++ {
+		dp[i] = dp[i-1] + nums[i-1]
+	}
+	return dp[1:]
+}
+
+```
diff --git a/leetcode/1512.Number-of-Good-Pairs/1512. Number of Good Pairs.go b/leetcode/1512.Number-of-Good-Pairs/1512. Number of Good Pairs.go
@@ -0,0 +1,13 @@
+package leetcode
+
+func numIdenticalPairs(nums []int) int {
+	total := 0
+	for x := 0; x < len(nums); x++ {
+		for y := x + 1; y < len(nums); y++ {
+			if nums[x] == nums[y] {
+				total++
+			}
+		}
+	}
+	return total
+}
diff --git a/leetcode/1512.Number-of-Good-Pairs/1512. Number of Good Pairs_test.go b/leetcode/1512.Number-of-Good-Pairs/1512. Number of Good Pairs_test.go
@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1512 struct {
+	para1512
+	ans1512
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1512 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1512 struct {
+	one int
+}
+
+func Test_Problem1512(t *testing.T) {
+
+	qs := []question1512{
+
+		{
+			para1512{[]int{1, 2, 3, 1, 1, 3}},
+			ans1512{4},
+		},
+
+		{
+			para1512{[]int{1, 1, 1, 1}},
+			ans1512{6},
+		},
+
+		{
+			para1512{[]int{1, 2, 3}},
+			ans1512{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1512------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1512, q.para1512
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, numIdenticalPairs(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/1512.Number-of-Good-Pairs/README.md b/leetcode/1512.Number-of-Good-Pairs/README.md
@@ -0,0 +1,67 @@
+# [1512. Number of Good Pairs](https://leetcode.com/problems/number-of-good-pairs/)
+
+## 题目
+
+Given an array of integers `nums`.
+
+A pair `(i,j)` is called good if `nums[i] == nums[j]` and `i < j`.
+
+Return the number of good pairs.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,1,1,3]
+Output: 4
+Explanation: There are 4 good pairs (0,3), (0,4), (3,4), (2,5) 0-indexed.
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,1,1,1]
+Output: 6
+Explanation: Each pair in the array are good.
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [1,2,3]
+Output: 0
+
+```
+
+**Constraints**:
+
+- `1 <= nums.length <= 1000`
+- `1 <= nums[i] <= 100`
+
+## 题目大意
+
+给你一个整数数组 nums。如果一组数字 (i,j) 满足 nums[i] == nums[j] 且 i < j ，就可以认为这是一组好数对。返回好数对的数目。
+
+## 解题思路
+
+- 简单题，按照题目中好数对的定义，循环遍历判断两数是否相等，累加计数即可。
+
+## 代码
+
+```go
+package leetcode
+
+func numIdenticalPairs(nums []int) int {
+	total := 0
+	for x := 0; x < len(nums); x++ {
+		for y := x + 1; y < len(nums); y++ {
+			if nums[x] == nums[y] {
+				total++
+			}
+		}
+	}
+	return total
+}
+
+```
diff --git a/website/content/ChapterFour/1480.Running-Sum-of-1d-Array.md b/website/content/ChapterFour/1480.Running-Sum-of-1d-Array.md
@@ -0,0 +1,64 @@
+# [1480. Running Sum of 1d Array](https://leetcode.com/problems/running-sum-of-1d-array/)
+
+## 题目
+
+Given an array `nums`. We define a running sum of an array as `runningSum[i] = sum(nums[0]…nums[i])`.
+
+Return the running sum of `nums`.
+
+**Example 1**:
+
+```
+Input: nums = [1,2,3,4]
+Output: [1,3,6,10]
+Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
+
+```
+
+**Example 2**:
+
+```
+Input: nums = [1,1,1,1,1]
+Output: [1,2,3,4,5]
+Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
+
+```
+
+**Example 3**:
+
+```
+Input: nums = [3,1,2,10,1]
+Output: [3,4,6,16,17]
+
+```
+
+
+**Constraints**:
+
+- `1 <= nums.length <= 1000`
+- `-10^6 <= nums[i] <= 10^6`
+
+## 题目大意
+
+给你一个数组 nums 。数组「动态和」的计算公式为：runningSum[i] = sum(nums[0]…nums[i]) 。请返回 nums 的动态和。
+
+
+## 解题思路
+
+- 简单题，按照题意依次循环计算前缀和即可。
+
+## 代码
+
+```go
+package leetcode
+
+func runningSum(nums []int) []int {
+	dp := make([]int, len(nums)+1)
+	dp[0] = 0
+	for i := 1; i <= len(nums); i++ {
+		dp[i] = dp[i-1] + nums[i-1]
+	}
+	return dp[1:]
+}
+
+```
diff --git a/website/content/ChapterFour/1512.Number-of-Good-Pairs.md b/website/content/ChapterFour/1512.Number-of-Good-Pairs.md
diff --git a/website/content/menu/index.md b/website/content/menu/index.md
