Merge pull request doocs#138 from Neil94n/neil

update Solution

Author: yanglbme

solution/0004.Median of Two Sorted Arrays/Solution.js

@@ -0,0 +1,74 @@
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number}  012345
+ */
+var findMedianSortedArrays = function (nums1, nums2) {
+    if (nums1.length == 0 || nums2.length == 0) {
+        if ((nums1.length + nums2.length) % 2 == 1) {
+            const index = parseInt((nums1.length + nums2.length) / 2)
+            return nums2.length == 0 ? nums1[index] : nums2[index]
+        } else {
+            let nums = nums2.length == 0 ? nums1 : nums2
+            const index = nums.length / 2
+            return (nums[index - 1] + nums[index]) / 2
+        }
+    }
+
+    if (nums1.length > nums2.length) {
+        swap(nums1, nums2)
+    }
+    const M = nums1.length, N = nums2.length
+    let min = 0, max = M, half = parseInt((M + N + 1) / 2) // 连个数组合并的中间值
+    while (min <= max) {
+        let i = parseInt((min + max) / 2) // nums1 的索引值
+        let j = half - i // num2 的索引值
+        if (i < max && nums2[j - 1] > nums1[i]) {
+            min++
+        } else if (i > min && nums1[i - 1] > nums2[j]) {
+            max--
+        } else {
+            let maxLeft = 0
+            if (i == 0) {
+                maxLeft = nums2[j - 1]
+            } else if (j == 0) {
+                maxLeft = nums1[i - 1]
+            } else {
+                maxLeft = Math.max(nums1[i - 1], nums2[j - 1])
+            }
+            if ((M + N) % 2 == 1) {
+                return maxLeft
+            }
+            let minRight = 0
+            if (i == M) {
+                minRight = nums2[j]
+            } else if (j == N) {
+                minRight = nums1[i]
+            } else {
+                minRight = Math.min(nums1[i], nums2[j])
+            }
+            return (maxLeft + minRight) / 2
+        }
+    }
+    return 0
+};
+
+function swap(a, b) {
+    let tmp = a
+    a = b
+    b = tmp
+}
+
+const nums1 = [4, 5]
+const nums2 = [1, 2, 3]
+findMedianSortedArrays(nums1, nums2)
+
+/**
+ * 实现思路
+ * 先排除空数组的情况
+ * 数组从小到大排序
+ * 取小数组的中间值
+ * 取大数组的索引 = 总中间值-小数组中间值
+ * 循环直到符合条件
+ * 如果都不符合条件，那么说明中间值在两个数组的左边或者右边
+ */

solution/0005.Longest Palindromic Substring/Solution.js

@@ -0,0 +1,36 @@
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var longestPalindrome = function (s) {
+    let maxLength = 0, left = 0, right = 0
+    for (let i = 0; i < s.length; i++) {
+        let singleCharLength = getPalLenByCenterChar(s, i, i)
+        let doubleCharLength = getPalLenByCenterChar(s, i, i + 1)
+        let max = Math.max(singleCharLength, doubleCharLength)
+        if (max > maxLength) {
+            maxLength = max
+            left = i - parseInt((max - 1) / 2)
+            right = i + parseInt(max / 2)
+        }
+    }
+    return s.slice(left, right + 1)
+};
+
+function getPalLenByCenterChar(s, left, right) {
+    // 中间值为两个字符，确保两个字符相等
+    if (s[left] != s[right]){
+        return right - left // 不相等返回为1个字符串
+    }
+    while (left > 0 && right < s.length - 1) {
+        // 先加减再判断
+        left--
+        right++
+        if (s[left] != s[right]){
+            return right - left - 1
+        }
+    }
+    return right - left + 1
+}
+
+console.log(longestPalindrome("cbbd"))