docs: 新增算法解题

Author: zjgyb

docs/views/algorithm/2022-04-08.md

@@ -11,6 +11,214 @@ tags:
 
 之前的[算法打卡](https://zjgyb.github.io/views/algorithm/2021-5-30.html)已经快一年了，由于代码行数较多，因此打开缓慢，故决定再新建一个，用于下一周期的算法打卡
 
+## 【贪心算法】盛最多水的容器
+[地址](https://leetcode.cn/problems/container-with-most-water/)
+
+```ts
+// 从左到右，从右到左，分别计算
+function maxArea(height: number[]): number {
+    const len = height.length
+    let result = 0
+    for (let i = 0; i < len; i++) {
+        let last = len - 1
+        while (last > i && height[last] < height[i]) {
+            last--
+        }
+
+        result = Math.max(result, (last - i) * height[i])
+    }
+
+    for (let i = len - 1; i >= 0; i--) {
+        let last = 0
+        while (last < i && height[last] < height[i]) {
+            last++
+        }
+
+        result = Math.max(result, (i - last) * height[i])
+    }
+
+    return result
+};
+```
+
+【双指针】：从最左和最右开始，较小的值移动
+
+```ts
+function maxArea(height: number[]): number {
+    let result = 0
+    let prev = 0
+    let next = height.length - 1
+    while (prev < next) {
+        result = Math.max(result, Math.min(height[prev], height[next]) * (next - prev))
+        height[prev] > height[next] ? next-- : prev++
+    }
+    return result
+};
+```
+
+
+## 【字符串】最简分数
+[地址](https://leetcode.cn/problems/simplified-fractions/)
+```ts
+function simplifiedFractions(n: number): string[] {
+    const result: string[] = []
+    let prev = 1
+
+    // 判断是否存在最小公倍数
+    const isPair = (prev: number, next: number) => {
+        const half = Math.min(Math.floor(next / 2), prev)
+        let start = 2
+        while(start <= half) {
+            if (prev % start === 0 && next % start === 0) {
+                return true
+            }
+            start++
+        }
+        return false
+    }
+
+
+    while(prev <= n) {
+        let next = prev + 1
+        while(next <= n) {
+            if (prev === 1 || !isPair(prev, next)) {
+                result.push(`${prev}/${next}`)
+            }
+            next++
+        }
+        prev++
+    }
+
+    return result
+};
+```
+
+```ts
+function simplifiedFractions(n: number): string[] {
+    const result: string[] = []
+    const set = new Set()
+    let prev = 1
+
+    while(prev <= n) {
+        let next = prev + 1
+        while(next <= n) {
+            if (prev === 1 || !set.has(`${prev},${next}`)) {
+                result.push(`${prev}/${next}`)
+                let i = 2
+
+                // 存起来
+                while(next * i <= n) {
+                    set.add(`${prev * i},${next * i}`)
+                    i++
+                }
+            }
+            next++
+        }
+        prev++
+    }
+
+    return result
+};
+```
+
+## 【字符串】 从英文中重建数字
+[地址](https://leetcode.cn/problems/reconstruct-original-digits-from-english/)
+
+```ts
+// 找到特征字符
+function originalDigits(s: string): string {
+    const obj = {
+        'zero': 0,
+        'xis': 6,
+        'seven': 7,
+        'wto': 2,
+        'geiht': 8,
+        'vfie': 5,
+        'ufor': 4,
+        'rthee': 3,
+        'one': 1,
+        'inne': 9
+    }
+    const words = {}
+    for (const word of s) {
+        words[word] = (words[word] || 0) + 1
+    }
+
+    const arr: number[] = Array(10).fill(0);
+    Object.keys(obj).forEach((key) => {
+        if (words[key[0]]) {
+            let num = words[key[0]]
+            arr[obj[key]] = num
+            for (const k of key) {
+                words[k] -= num
+            }
+        }
+    })
+
+    return arr.reduce((prev, next, index) => {
+        if (next !== 0) {
+            prev += `${index}`.repeat(next)
+        }
+        return prev
+    }, '')
+};
+```
+
+## 【字符串】字母异位词分组
+[地址](https://leetcode.cn/problems/group-anagrams/)
+
+```ts
+function groupAnagrams(strs: string[]): string[][] {
+    const results: string[][] = []
+    const map = new Map()
+
+    for (const str of strs) {
+        // 切割后排序
+        const result = str.split('').sort((a, b) => a.codePointAt(0) - b.codePointAt(0)).join('')
+
+        if (map.has(result)) {
+            results[map.get(result)].push(str)
+        } else {
+            map.set(result, results.length)
+            results[results.length] = [str]
+        }
+    }
+
+    return results
+};
+```
+
+## 【链表】合并零之间的节点
+[地址](https://leetcode.cn/problems/merge-nodes-in-between-zeros/)
+
+```ts
+// 思路就是快慢指针
+function mergeNodes(head: ListNode | null): ListNode | null {
+    if (!head) {
+        return head
+    }
+
+    let fast = head
+    let slow = head
+    let val = 0
+
+    while (fast?.next) {
+        val += fast.val
+        if (fast.next.val !== 0) {
+            fast = fast.next
+            continue
+        }
+
+        slow.next = fast.next
+        fast.next.val = val
+        val = 0
+        slow = slow.next
+        fast = slow.next
+    }
+    return head.next
+};
+```
+
 ## 【字符串】检查相同字母间的距离
 [地址](https://leetcode.cn/problems/check-distances-between-same-letters/)