Majority Element

Change-Id: I57339ea91629b95bd318d8df6b126fcb2d911f1c

Author: applewjg

MajorityElement.h

@@ -0,0 +1,36 @@
+/*
+ Author:     King, [email protected]
+ Date:       Dec 14, 2014
+ Problem:    ZigZag Conversion
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/majority-element/
+ Notes:
+ Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
+
+ You may assume that the array is non-empty and the majority element always exist in the array.
+
+ Solution: Runtime: O(n) — Moore voting algorithm: We maintain a current candidate and a counter initialized to 0. As we iterate the array, we look at the current element x:
+ If the counter is 0, we set the current candidate to x and the counter to 1.
+ If the counter is not 0, we increment or decrement the counter based on whether x is the current candidate.
+ After one pass, the current candidate is the majority element. Runtime complexity = O(n).
+ */
+
+class Solution {
+public:
+    int majorityElement(vector<int> &num) {
+        int n = num.size();
+        if (n == 0) return 0;
+        if (n == 1) return num[0];
+        int cur = num[0], cnt = 1;
+        for (int i = 1; i < num.size(); ++i) {
+            if (cnt == 0) {
+                cur = num[i];
+                ++cnt;
+                continue;
+            }
+            if (cur == num[i]) ++cnt;
+            else --cnt;
+        }
+        return cur;
+    }
+};