规范格式

leetcode/0027.Remove-Element/27. Remove Element.go

@@ -0,0 +1,19 @@
+package leetcode
+
+func removeElement(nums []int, val int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	j := 0
+	for i := 0; i < len(nums); i++ {
+		if nums[i] != val {
+			if i != j {
+				nums[i], nums[j] = nums[j], nums[i]
+				j++
+			} else {
+				j++
+			}
+		}
+	}
+	return j
+}

leetcode/0027.Remove-Element/27. Remove Element_test.go

@@ -0,0 +1,68 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question27 struct {
+	para27
+	ans27
+}
+
+// para 是参数
+// one 代表第一个参数
+type para27 struct {
+	one []int
+	two int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans27 struct {
+	one int
+}
+
+func Test_Problem27(t *testing.T) {
+
+	qs := []question27{
+
+		question27{
+			para27{[]int{1, 0, 1}, 1},
+			ans27{1},
+		},
+
+		question27{
+			para27{[]int{0, 1, 0, 3, 0, 12}, 0},
+			ans27{3},
+		},
+
+		question27{
+			para27{[]int{0, 1, 0, 3, 0, 0, 0, 0, 1, 12}, 0},
+			ans27{4},
+		},
+
+		question27{
+			para27{[]int{0, 0, 0, 0, 0}, 0},
+			ans27{0},
+		},
+
+		question27{
+			para27{[]int{1}, 1},
+			ans27{0},
+		},
+
+		question27{
+			para27{[]int{0, 1, 2, 2, 3, 0, 4, 2}, 2},
+			ans27{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 27------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans27, q.para27
+		fmt.Printf("【input】:%v    【output】:%v\n", p.one, removeElement(p.one, p.two))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0027.Remove-Element/README.md

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+# [27. Remove Element](https://leetcode.com/problems/remove-element/)
+
+## 题目
+
+Given an array nums and a value val, remove all instances of that value in-place and return the new length.
+
+Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
+
+The order of elements can be changed. It doesn't matter what you leave beyond the new length.
+
+Example 1:
+
+```c
+Given nums = [3,2,2,3], val = 3,
+
+Your function should return length = 2, with the first two elements of nums being 2.
+
+It doesn't matter what you leave beyond the returned length.
+```
+
+Example 2:
+
+```c
+Given nums = [0,1,2,2,3,0,4,2], val = 2,
+
+Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.
+
+Note that the order of those five elements can be arbitrary.
+
+It doesn't matter what values are set beyond the returned length.
+```
+
+Clarification:
+
+Confused why the returned value is an integer but your answer is an array?
+
+Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
+
+Internally you can think of this:
+
+```c
+// nums is passed in by reference. (i.e., without making a copy)
+int len = removeElement(nums, val);
+
+// any modification to nums in your function would be known by the caller.
+// using the length returned by your function, it prints the first len elements.
+for (int i = 0; i < len; i++) {
+    print(nums[i]);
+}
+```
+
+## 题目大意
+
+给定一个数组 nums 和一个数值 val，将数组中所有等于 val 的元素删除，并返回剩余的元素个数。
+
+## 解题思路
+
+这道题和第 283 题很像。这道题和第 283 题基本一致，283 题是删除 0，这一题是给定的一个 val，实质是一样的。
+
+这里数组的删除并不是真的删除，只是将删除的元素移动到数组后面的空间内，然后返回数组实际剩余的元素个数，OJ 最终判断题目的时候会读取数组剩余个数的元素进行输出。