feat: udpate

04-algrithm/02-finger-guess/__tests__/index.spec.js

@@ -0,0 +1,20 @@
+const { check } = require('../index');
+/*少数服从多数的方式常在人们意见不一致时使用。该方式简单明了，所以除了政界，学校和公司也常
+结果
+当当一定数量的人猜拳时，一次就能定胜负的人数组合方式有多少种呢？以4人为例，表1.3列出了所有可能的组合方式，一共有 12 种。
+*/
+describe('猜拳少数服从多数', () => {
+    describe('递归解法', () => {
+        it('3人 9种情况', () => {
+            expect(check(3)).toBe(9);
+        })
+
+        it('4人 12种情况', () => {
+            expect(check(4)).toBe(12);
+        })
+
+        it('100人 5100种情况', () => {
+            expect(check(100)).toBe(5100);
+        })
+    })
+})

04-algrithm/02-finger-guess/index.js

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+// finger guess
+exports.check = function check(n) {
+    var cnt = 0;
+    for (var rock = 0; rock <= n; rock++) {
+        for (var paper = 0; paper <= n - rock; paper++) {
+            var scissors = n - rock - paper;
+            if (rock > scissors) {
+                if (rock != paper) {
+                    cnt++;
+                }
+            } else if (rock < scissors) {
+                if (paper != scissors) {
+                    cnt++;
+                }
+            } else {
+                // 相等的情况
+                if (paper > rock) {
+                    cnt++;
+                }
+            }
+        }
+    }
+    return cnt;
+}

04-algrithm/03-light/__tests__/index.spec.js

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+const { getNum } = require('../index')
+// 24小时制
+describe('电子钟的亮灯数', () => {
+    test('27处亮灯 8800种', () => {
+        expect(getNum(27)).toBe(8800)
+    })
+
+    test('30处亮灯 8360种', () => {
+        expect(getNum(30)).toBe(8360)
+    })
+})

04-algrithm/03-light/index.js

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+exports.getNum = function (N) {
+    /**
+     * 获取两位数的亮灯数
+     */
+    function check(num) {
+        var light = [6, 2, 5, 5, 4, 5, 6, 3, 7, 6]
+        return light[Math.floor(num / 10)] + light[num % 10]
+    }
+
+    var lights = new Array(60)
+    for (var i = 0; i < 60; i++) {
+        lights[i] = check(i)
+    }
+
+    var cnt = 0
+    for (var h = 0; h < 24; h++) {
+        for (var m = 0; m < 60; m++) {
+            for (var s = 0; s < 60; s++) {
+                if (lights[h] + lights[m] + lights[s] === N) {
+                    cnt++
+                }
+            }
+        }
+    }
+    return cnt
+}

04-algrithm/04-fibonacci/__tests__/index.spec.js

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+const { fi, fi2 } = require('../index')
+
+describe('斐波那契数列', () => {
+    describe('递归', () => {
+        it('fibonacci(9) => 34', () => {
+            expect(fi(9)).toBe(55)
+        })
+        it('fibonacci(16) => 987', () => {
+            expect(fi(40)).toBe(165580141)
+        })
+    })
+
+    describe('动态规划', () => {
+        it('fibonacci(9) => 34', () => {
+            expect(fi2(9)).toBe(55)
+        })
+        it('fibonacci(16) => 987', () => {
+            expect(fi2(40)).toBe(165580141)
+        })
+    })
+
+
+})

04-algrithm/04-fibonacci/index.js

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+exports.fi = function fi(n) {
+    if (n === 0) return 1
+    if (n === 1) return 1
+    if (n === 2) return 2
+    return fi(n - 1) + fi(n - 2)
+}
+
+exports.fi2 = function fi2(n) {
+    if (n === 0) return 1
+    if (n === 1) return 1
+    if (n === 2) return 2
+    var a = 1
+    var b = 1
+    var c = 2
+    for (var i = 3; i <= n; i++) {
+        a = b
+        b = c
+        c = a + b
+    }
+    return c
+}
+
+
+// for (var i = 0; i < 10; i++) {
+//     console.log(fi2(i))
+// }
+

04-algrithm/05-coin/__tests__/index.spec.js

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+const { check } = require('../index')
+describe('杨辉三角', () => {
+    it('第1层 需要的2个硬币数', () => {
+        expect(check(1)).toBe(2)
+    })
+    it('第2层 需要的8个硬币数', () => {
+        expect(check(3)).toBe(8)
+    })
+    it('第45层 需要的3518437540个硬币数', () => {
+        expect(check(45)).toBe(3518437540)
+    })
+})

04-algrithm/05-coin/index.js

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+exports.check = function (n) {
+    function getCount(n) {
+        var coin = [10000, 5000, 2000, 1000, 500, 100, 50, 10, 5, 1]
+        result = 0
+        for (var i = 0; i < coin.length; i++) {
+            var cnt = Math.floor(n / coin[i])
+            n = n % coin[i]
+            result += cnt
+        }
+        return result
+    }
+    row = new Array(n + 1)
+    row[0] = 1
+    for (var i = 1; i < n + 1; i++) {
+        row[i] = 0
+    }
+    for (var i = 0; i < n; i++) {
+        for (var j = i + 1; j > 0; j--) {
+            row[j] += row[j - 1]
+        }
+    }
+    var total = 0
+    for (var i = 0; i < n + 1; i++) {
+        total += getCount(row[i])
+    }
+    return total
+}