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| 1 | +////////////////////////////////////////////////////////////////////////////// |
| 2 | +// Two Stacks |
| 3 | +// Time: O(1) |
| 4 | +// Space: O(n) |
| 5 | +// This solution uses two stacks to save the total values and the minimum |
| 6 | +// values. Per each new value if it is less than or equal to the current |
| 7 | +// minimum it is pushed to both stacks. We save duplicate minimum values to |
| 8 | +// avoid the conundrum of inquiring whether the minimum value can be removed |
| 9 | +// (i.e. when the minimum value equals the top value can it be removed or are |
| 10 | +// there duplicate values in the main stack). |
| 11 | +////////////////////////////////////////////////////////////////////////////// |
| 12 | + |
| 13 | +classMinStack{ |
| 14 | +/** |
| 15 | + *@constructor |
| 16 | + */ |
| 17 | +constructor(){ |
| 18 | +this.mainStack=[]; |
| 19 | +this.minStack=[]; |
| 20 | +} |
| 21 | + |
| 22 | +/** |
| 23 | + *@param {number} val |
| 24 | + *@return {void} |
| 25 | + */ |
| 26 | +push(val){ |
| 27 | +this.mainStack.push(val); |
| 28 | +if(!this.minStack.length||val<=this.minStack[this.minStack.length-1]){ |
| 29 | +this.minStack.push(val); |
| 30 | +} |
| 31 | +} |
| 32 | + |
| 33 | +/** |
| 34 | + *@return {void} |
| 35 | + */ |
| 36 | +pop(){ |
| 37 | +constval=this.mainStack.pop(); |
| 38 | +if(val===this.minStack[this.minStack.length-1]){ |
| 39 | +this.minStack.pop(); |
| 40 | +} |
| 41 | +} |
| 42 | + |
| 43 | +/** |
| 44 | + *@return {number} |
| 45 | + */ |
| 46 | +top(){ |
| 47 | +returnthis.mainStack[this.mainStack.length-1]; |
| 48 | +} |
| 49 | + |
| 50 | +/** |
| 51 | + *@return {number} |
| 52 | + */ |
| 53 | +getMin(){ |
| 54 | +returnthis.minStack[this.minStack.length-1]; |
| 55 | +} |
| 56 | +} |