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| 1 | +/** |
| 2 | + * @see {@link https://leetcode.com/problems/merge-two-sorted-lists/?envType=study-plan&id=level-1 21. Merge Two Sorted Lists} |
| 3 | + */ |
| 4 | + |
| 5 | +/** |
| 6 | + * EXPLAIN |
| 7 | + * |
| 8 | + * given two LL where vals of each node are guaranteed to be the lowest in each LL |
| 9 | + * need to return the HEAD of each list |
| 10 | + * not guaranteed to be same list size |
| 11 | + */ |
| 12 | + |
| 13 | +/** |
| 14 | + * APPROACH |
| 15 | + * |
| 16 | + * create new LL |
| 17 | + * create separate ref value to new LL |
| 18 | + * create refs to L1 and L2 |
| 19 | + * |
| 20 | + * WHILE LOOP until both LL dont exist |
| 21 | + * |
| 22 | + * IF L1 DNE but L2 exists |
| 23 | + * create next on new LL using L2 |
| 24 | + * advance ref on new LL |
| 25 | + * continue |
| 26 | + * IF L2 but DNE but L1 exists |
| 27 | + * create next of new LL using L1 |
| 28 | + * advance ref on new LL |
| 29 | + * continue |
| 30 | + * |
| 31 | + * IF L1 val greater than L2 |
| 32 | + * create next on new LL using L1 val |
| 33 | + * advance ref to L1 next |
| 34 | + * advance ref on new LL |
| 35 | + * ELSE IF L2 val greater than L1 |
| 36 | + * create next on new LL using L2 val |
| 37 | + * advance ref to L2 next |
| 38 | + * advance ref on new LL |
| 39 | + * ELSE both L1 and L2 are the same |
| 40 | + * create next on new LL using L1 |
| 41 | + * create next.next on new LL using L2 |
| 42 | + * advance both L1 and L2 to next |
| 43 | + * advance new LL twice |
| 44 | + * LOOP ENDS when L1 and L2 DNE |
| 45 | + * |
| 46 | + * return separate ref val to new LL |
| 47 | + */ |
| 48 | + |
| 49 | +/** |
| 50 | + * CODE [FAIL] |
| 51 | + * issue where I would get leading 0 value for undefined vals when handling empty lists |
| 52 | + */ |
| 53 | +/** |
| 54 | + * Definition for singly-linked list. |
| 55 | + * function ListNode(val, next) { |
| 56 | + * this.val = (val===undefined ? 0 : val) |
| 57 | + * this.next = (next===undefined ? null : next) |
| 58 | + * } |
| 59 | + */ |
| 60 | +/** |
| 61 | + * @param {ListNode} list1 |
| 62 | + * @param {ListNode} list2 |
| 63 | + * @return {ListNode} |
| 64 | + */ |
| 65 | +var mergeTwoLists = function (list1, list2) { |
| 66 | + const merged = new ListNode(); |
| 67 | + let mergedRef = merged; |
| 68 | + |
| 69 | + // BASE CASE: NO NODES |
| 70 | + if (!list1 && !list2) return new ListNode(); |
| 71 | + |
| 72 | + while (list1 || list2) { |
| 73 | + // CASES: one of the lists in null, other still has vals |
| 74 | + if (!list1) { |
| 75 | + mergedRef.next = new ListNode(list2.val); |
| 76 | + mergedRef = mergedRef.next; |
| 77 | + list2 = list2.next; |
| 78 | + continue; |
| 79 | + } |
| 80 | + |
| 81 | + if (!list2) { |
| 82 | + mergedRef.next = new ListNode(list1.val); |
| 83 | + mergedRef = mergedRef.next; |
| 84 | + list1 = list1.next; |
| 85 | + continue; |
| 86 | + } |
| 87 | + |
| 88 | + // CASES: both lists still have vals |
| 89 | + if (list1.val > list2.val) { |
| 90 | + mergedRef.next = new ListNode(list1.val); |
| 91 | + mergedRef = mergedRef.next; |
| 92 | + list1 = list1.next; |
| 93 | + } else if (list2.val > list1.val) { |
| 94 | + mergedRef.next = new ListNode(list2.val); |
| 95 | + mergedRef.next = mergedRef; |
| 96 | + list2 = list2.next; |
| 97 | + } else { |
| 98 | + mergedRef.next = new ListNode(list1.val); |
| 99 | + mergedRef.next.next = new ListNode(list2.val); |
| 100 | + mergedRef = mergedRef.next.next; |
| 101 | + list1 = list1.next; |
| 102 | + list2 = list2.next; |
| 103 | + } |
| 104 | + } |
| 105 | + return merged; |
| 106 | +}; |
| 107 | + |
| 108 | +/** |
| 109 | + * Compare with working solution |
| 110 | + */ |
| 111 | +/** |
| 112 | + * Definition for singly-linked list. |
| 113 | + * function ListNode(val, next) { |
| 114 | + * this.val = (val===undefined ? 0 : val) |
| 115 | + * this.next = (next===undefined ? null : next) |
| 116 | + * } |
| 117 | + */ |
| 118 | +/** |
| 119 | + * @param {ListNode} list1 |
| 120 | + * @param {ListNode} list2 |
| 121 | + * @return {ListNode} |
| 122 | + */ |
| 123 | +var mergeTwoLists = function (list1, list2) { |
| 124 | + let head = new ListNode(), |
| 125 | + tail; |
| 126 | + |
| 127 | + if (!list1 || !list2) { |
| 128 | + return list2 || list1; |
| 129 | + } |
| 130 | + |
| 131 | + if (list1.val < list2.val) { |
| 132 | + head = list1; |
| 133 | + list1 = list1.next; |
| 134 | + } else { |
| 135 | + head = list2; |
| 136 | + list2 = list2.next; |
| 137 | + } |
| 138 | + tail = head; |
| 139 | + |
| 140 | + while (list1 && list2) { |
| 141 | + if (list1.val < list2.val) { |
| 142 | + tail.next = list1; |
| 143 | + tail = tail.next; |
| 144 | + list1 = list1.next; |
| 145 | + } else if (list1.val > list2.val) { |
| 146 | + tail.next = list2; |
| 147 | + tail = tail.next; |
| 148 | + list2 = list2.next; |
| 149 | + } else if (list1.val == list2.val) { |
| 150 | + tail.next = list1; |
| 151 | + tail = tail.next; |
| 152 | + list1 = list1.next; |
| 153 | + tail.next = list2; |
| 154 | + tail = tail.next; |
| 155 | + list2 = list2.next; |
| 156 | + } |
| 157 | + } |
| 158 | + tail.next = list1 || list2; |
| 159 | + return head; |
| 160 | +}; |
| 161 | + |
| 162 | +/** |
| 163 | + * SOLUTION USED |
| 164 | + * Recursive solution |
| 165 | + * https://leetcode.com/problems/merge-two-sorted-lists/solutions/2705782/js-recursion-with-exlanation/?envType=study-plan&id=level-1&orderBy=most_votes&languageTags=javascript |
| 166 | + */ |
| 167 | + |
| 168 | +var mergeTwoLists = function (l1, l2) { |
| 169 | + if (!l1) return l2; |
| 170 | + else if (!l2) return l1; |
| 171 | + else if (l1.val <= l2.val) { |
| 172 | + l1.next = mergeTwoLists(l1.next, l2); |
| 173 | + return l1; |
| 174 | + } else { |
| 175 | + l2.next = mergeTwoLists(l1, l2.next); |
| 176 | + return l2; |
| 177 | + } |
| 178 | +}; |
| 179 | + |
| 180 | +/** |
| 181 | + * WHILE LOOP Solution |
| 182 | + * https://leetcode.com/problems/merge-two-sorted-lists/solutions/2630303/fast-typescript-solution-75-ms-top-95-speed/ |
| 183 | + */ |
| 184 | +function mergeTwoLists( |
| 185 | + list1: ListNode | null, |
| 186 | + list2: ListNode | null |
| 187 | +): ListNode | null { |
| 188 | + const mergedHead: ListNode = new ListNode(-1, null); |
| 189 | + let prev: ListNode = mergedHead; |
| 190 | + |
| 191 | + while (list1 && list2) { |
| 192 | + if (list1.val <= list2.val) { |
| 193 | + prev.next = list1; |
| 194 | + list1 = list1.next; |
| 195 | + } else { |
| 196 | + prev.next = list2; |
| 197 | + list2 = list2.next; |
| 198 | + } |
| 199 | + prev = prev.next; |
| 200 | + } |
| 201 | + prev.next = list1 ? list1 : list2; |
| 202 | + |
| 203 | + return mergedHead.next; |
| 204 | +} |
| 205 | + |
| 206 | +/** |
| 207 | + * READ 09:08 |
| 208 | + * EXPLAIN 06:51 15:59 |
| 209 | + * APPROACH 05:08 21:08 |
| 210 | + * CODE 09:01 30:09 |
| 211 | + * TEST [FAIL] 09:44 39:53 |
| 212 | + * OPTIMIZE |
| 213 | + */ |
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